SlideShare a Scribd company logo

Wave Optics Interference

β€’
β€’
R
RahulGupta29181

The document discusses wave optics and the wave nature of light. It explains that light behaves as waves under certain conditions and can exhibit phenomena like interference and diffraction. It describes key concepts in wave optics like wavefronts, Huygens' principle, interference, and Young's double slit experiment. The double slit experiment demonstrates the wave nature of light by producing an interference pattern on a screen from a coherent light source passing through two slits.

Science
1 of 121
Download to read offline
Welcome to Wave optics
Does Light always exhibit Rectilinear Propagation?
Explanation for bending of light Wave theory of light Christiaan Huygens Wavefront Huygens’ Principle Huygens Wave theory of light
ο‚— All points on a particular circle will oscillate with the same phase. ο‚— Wavefront is the locus of all points at which the wave disturbance is in the same phase. Wavefront ο‚— In 3𝐷, for a point source, the wavefronts are spherically symmetric. ο‚— Point source creates wave traveling in all directions according to equation. 𝑦 = 𝐴sin(πœ”π‘‘ Β± π‘˜π‘₯) ο‚— Points 𝐴, 𝐡, 𝐢 have same phase πœ”π‘‘ Β± π‘˜π‘₯1.
The direction of propagation of light wave is perpendicular to the wavefront at any given point. Direction of Propagation of Light Spherical Wavefront
ο‚— Time taken by each ray to propagate from one wavefront to the next wavefront in any medium remains same. ο‚— Hence, distance covered by the light in each time interval also changes. ⟹ Speed of light changes. ο‚— Refractive index, 𝑛 = 𝑐 𝑣 When medium changes, 𝑛1 𝑛2 = 𝑣2 𝑣1 Properties of Wavefront πœ† ο‚— Distance between two consecutive wavefronts = πœ†.
ο‚— Spherical wavefront is observed for a point source. β€’ Direction of propagation of light: Radially outward and perpendicular to the wavefront at any given point. Spherical Wavefront
β€’ Planar wavefronts are observed when the source is at infinity. β€’ Direction of propagation of light: Perpendicular to the plane. Planar Wavefront
β€’ Cylindrical wavefront is observed for a line source. β€’ Direction of propagation of light: Radially outward and perpendicular to the wavefront at any given point. Cylindrical Wavefront
S If the distance between the wavefronts in the medium with refractive index 𝑛2 is 𝑑2, then what will be the distance 𝑑1 between the wavefronts in the medium with refractive index 𝑛1? 𝑛 = 𝑐 𝑣 We known that, ΰ΅— 𝑛2 𝑛1 = ΰ΅— 𝑣1 𝑣2 πœ† = ΰ΅— 𝑣 𝑓 β‡’ β‡’ ΰ΅— πœ†1 πœ†2 = ΰ΅— 𝑣1 𝑣2 = ΰ΅— 𝑛2 𝑛1 Here, πœ†1 = 𝑑1, and πœ†2 = 𝑑2 Hence, 𝑑1 𝑑2 = 𝑛2 𝑛1 β‡’ 𝑑1 = 𝑑2 𝑛2 𝑛1
ο‚— Every point on the wavefront acts as a point source called secondary wave source and generates secondary wavelets. ο‚— The common tangent to the secondary wavelets in the forward direction gives the secondary wavefront. ο‚— Intensity is maximum in forward direction and zero in backward direction. Secondary Wavelets Secondary Wavefront Primary Wavefront Primary Source Huygens’ Principle
Shape of Wavefront Object at infinity for a convex lens Object at infinity for a concave lens
Object at infinity for a concave mirror Shape of Wavefront Object at infinity for a prism
Reflection of a plane wave by a plane surface Huygens’ Principle - Law of Reflection Reflecting plane surface 𝐴 𝐡 𝐡𝐢 = 𝑣𝑑 𝐢 𝐷 𝐷𝐴 = 𝑣𝑑 𝑑 = 0 𝑖 𝑑 Incident Wavefront 𝐴𝐡 at 𝑑 = 0 Reflected Wavefront 𝐢𝐷 at time = 𝑑 Secondary Wavelet From 𝐴
Reflection of a plane wave by a plane surface Huygens’ Principle - Law of Reflection ∠𝐷𝐢𝐴 = π‘Ÿ β†’ Angle of reflection ∠𝐡𝐴𝐢 = 𝑖 β†’ Angle of incidence ο‚— Angle of incidence is the angle between the incident wavefront and the reflecting surface. ο‚— Angle of reflection is the angle between the reflected wavefront and the reflecting surface. 𝐴 𝐢 𝐷 π‘Ÿ 𝐡 𝐢 𝐴 𝑖 ∴ βˆ†π΄π·πΆ β‰… βˆ†πΆπ΅π΄ ⟹ βˆ π‘– = βˆ π‘Ÿ From βˆ†π΄π·πΆ and βˆ†πΆπ΅π΄: ∠𝐴𝐷𝐢 = ∠𝐴𝐡𝐢 = 90Β° 𝐴𝐷 = 𝐡𝐢 = 𝑣𝑑 𝐴𝐢 = 𝐴𝐢 [𝑅. 𝐻. 𝑆 Congruency]
From βˆ†π΄π·πΆ: From βˆ†π΄π΅πΆ: sin 𝑖 = 𝐡𝐢 𝐴𝐢 = 𝑣1 𝑑 𝐴𝐢 sin π‘Ÿ = 𝐴𝐷 𝐴𝐢 = 𝑣2 𝑑 𝐴𝐢 sin 𝑖 sin π‘Ÿ = 𝑣1 𝑣2 = 𝑛2 𝑛1 = Constant Huygens’ Principle - Law of Refraction Medium 𝑛2 𝐴 𝐡 𝑣1𝑑 𝐢 𝐷 𝑖 π‘Ÿ 𝑑 = 0 π‘Ÿ 𝑖 Incident Wavefront 𝐴𝐡 at 𝑑 = 0 Refracted Wavefront 𝐢𝐷 at 𝑑 = 𝑑 Secondary Wavelet From A 𝑣2𝑑 π‘Ÿ Medium 𝑛1 𝑛1 < 𝑛2
B A D C Solution: The direction of propagation β†’ ΖΈ 𝑖 ⟹ The direction of wavefront β†’ βŠ₯ π‘‘π‘œ ΖΈ 𝑖 In the given options, the plane βŠ₯ to ΖΈ 𝑖 is represented by π‘₯ = 𝑐. π‘₯ = 𝑐 𝑦 = 𝑐 𝑧 = 𝑐 π‘₯ + 𝑦 + 𝑧 = 𝑐 Light waves travel in vacuum, along the π‘₯ βˆ’ axis. Which of the following may represent the wavefronts?
Hole Screen Hole Screen Light behaves like a ray (i.e., exhibits rectilinear propagation) for: Obstacle dimensions β‰«πœ†π‘™π‘–π‘”β„Žπ‘‘ Obstacle dimensions β‰ˆ πœ†π‘™π‘–π‘”β„Žπ‘‘ Light behaves like a wave for: Does Light always exhibit Rectilinear Propagation?
OPTICS Study of the behavior and properties of light GEOMETRICAL OPTICS Explains refraction and reflection, but not interference, diffraction and polarization WAVE OPTICS Explains reflection and refraction, as well as interference, diffraction and polarization
Interference is the phenomenon in which two waves superpose to form the resultant wave of lower, higher or same amplitude. Interference
β€œWhen two or more waves cross at a point, the displacement at that point is equal to the vector sum of the displacements of individual waves” Τ¦ 𝑦𝑛𝑒𝑑 = Τ¦ 𝑦1 + Τ¦ 𝑦2 + Τ¦ 𝑦3 … … + Τ¦ 𝑦𝑛 𝑦 π‘₯ 𝑂 𝑦1 𝑦2 𝑦𝑛𝑒𝑑 Superposition Principle
Case 1: When two crest meet Case 2: When crest and trough meet Resultant of Waves
Phase Difference and Path Difference ● Path difference (βˆ†π‘₯): Difference in the path traversed by the two waves. ● Phase difference (𝛿): Difference in the phase angle of the two waves. We know that, For a path difference πœ†, phase difference = 2πœ‹ So, for path difference βˆ†π‘₯, phase difference = 2πœ‹ πœ† βˆ†π‘₯. 𝛿 = 2πœ‹ πœ† βˆ†π‘₯ Source 𝑆1 Source 𝑆2 𝑦1 = 𝐴1 sin πœ”π‘‘ + π‘˜π‘₯ 𝑦2 = 𝐴2 sin πœ”π‘‘ + π‘˜π‘₯ + 𝛿 𝑦𝑛𝑒𝑑 = 𝑦1 + 𝑦2 𝑃
Combination of Waves 𝑦𝑛𝑒𝑑 = 𝐴1 sin πœ”π‘‘ + π‘˜π‘₯ + 𝐴2 sin πœ”π‘‘ + π‘˜π‘₯ + 𝛿 𝐴𝑛𝑒𝑑 = 𝐴1 2 + 𝐴2 2 + 2𝐴1𝐴2 cos 𝛿 tan 𝛼 = 𝐴2 sin 𝛿 𝐴1 + 𝐴2 cos 𝛿 𝑦𝑛𝑒𝑑 = 𝐴𝑛𝑒𝑑 sin πœ”π‘‘ + π‘˜π‘₯ + 𝛼 If 𝐴1 = 𝐴2 = 𝐴 𝐴𝑛𝑒𝑑 = 2𝐴 cos 𝛿 2
Constructive Interference Interference that produces maximum possible amplitude (or maximum intensity) is called constructive interference. 𝑃 2πœ† 3πœ† ● Path Difference = 3πœ† βˆ’ 2πœ† = πœ† 𝐴𝑛𝑒𝑑 = 𝐴1 2 + 𝐴2 2 + 2𝐴1𝐴2 cos 𝛿 For maximum amplitude: cos 𝛿 = 1 β‡’ 𝛿 = 2π‘›πœ‹ 𝐴 = π΄π‘šπ‘Žπ‘₯ = 𝐴1 + 𝐴2 2π‘›πœ‹ = 2πœ‹ πœ† Γ— (βˆ†π‘₯) Path difference = βˆ†π‘₯ = π‘›πœ† 𝛿 = 2πœ‹ πœ† βˆ†π‘₯ Constructive Interference
≑ π‘₯ 𝑦 𝑦𝑛𝑒𝑑 = 2𝐴 sin(πœ”π‘‘ + π‘˜π‘₯) + Constructive interference occurs when the crest and trough of one wave overlaps with the crest and trough of another wave. Constructive Interference π‘₯ 𝑦 𝑦1 = 𝐴 sin(πœ”π‘‘ + π‘˜π‘₯) 𝑂 π‘₯ 𝑦 𝑦2 = 𝐴 sin(πœ”π‘‘ + π‘˜π‘₯) 𝑂
Destructive Interference Interference that produces minimum possible amplitude (or minimum intensity) is called destructive interference. 𝐴𝑛𝑒𝑑 = 𝐴1 2 + 𝐴2 2 + 2𝐴1𝐴2 cos 𝛿 𝑃 7.25πœ† 9.75πœ† For minimum amplitude: cos 𝛿 = βˆ’1 β‡’ 𝛿 = (2𝑛 + 1)πœ‹ 𝐴 = π΄π‘šπ‘–π‘› = 𝐴1 βˆ’ 𝐴2 (2𝑛 + 1)πœ‹ = 2πœ‹ πœ† (βˆ†π‘₯) Path difference = 𝛿 = 2πœ‹ πœ† βˆ†π‘₯ βˆ†π‘₯ = 2𝑛 + 1 πœ† 2 ● Path Difference = 9.75πœ† βˆ’ 7.25πœ† = 2.5πœ† Destructive Interference
π‘₯ 𝑦 𝑦1 = 𝐴 sin(πœ”π‘‘ + π‘˜π‘₯) Destructive interference is resulted when the crest of one wave overlaps with the trough of another wave. ≑ π‘₯ 𝑦 𝑂 𝑦 = 0 + π‘₯ 𝑦 𝑦2 = 𝐴 sin(πœ”π‘‘ + π‘˜π‘₯ + πœ‹) Destructive Interference 𝑂 𝑂
T Two-point light sources 𝑆1 and 𝑆2 are separated by a distance of 4.2πœ†. If an observer standing at the centre 𝐢 of the two sources starts moving towards 𝑆2, then find the minimum distance travelled by the observer to meet the first maxima. 𝑆1 𝑆2 2.1πœ† 𝐢 𝑃 βˆ†π‘₯ = 0 π‘₯ 2.1πœ† βˆ†π‘₯ = πœ† Solution: For maxima at 𝑃: βˆ†π‘₯ = πœ† β‡’ 𝑆1𝑃 βˆ’ 𝑆2𝑃 = πœ† β‡’ (2.1πœ† + π‘₯) βˆ’ (2.1πœ† βˆ’ π‘₯) = πœ† β‡’ 2π‘₯ = πœ† π‘₯ = 0.5πœ†
Coherent and Incoherent Sources Coherent Waves Incoherent Waves Coherent Sources ● Same wavelength ● Same frequency ● Constant phase difference Incoherent Sources ● Different wavelength ● Different frequency ● Varying phase difference Note: We will always consider coherent sources in our discussion.
Young’s Double Slit Experiment Particle nature of light Wave nature of light
According to Huygens’ principle, the sources 𝑆1 and 𝑆2 will behave as independent sources. Young’s Double Slit Experiment 𝐷 ● Light source must be monochromatic. ● Sources 𝑆1 and 𝑆2 must be coherent. ● Width of the slit is comparable to the wavelength of light. ● Waves coming from sources 𝑆1 and 𝑆2 will interference and obtain different interference pattern on the screen.
Young’s Double Slit Experiment β€’ 𝑆𝑆1 = 𝑆𝑆2 𝐷 (∴ No path difference till slits.) 𝑑 β€’ ′𝑑′ is distance between slits. β€’ 𝑂𝐡 is the central line. 𝑂 𝐡 β€’ ′𝐷′ is distance between screen and slits plane. πœƒ 𝑃 𝑦 β€’ Consider point 𝑃 at a distance 𝑦 from central line 𝑂𝐡. β€’ βˆ π‘ƒπ‘‚π΅ = πœƒ. β€’ 𝑆2𝑃 > 𝑆1𝑃 β†’ 𝑆2𝑃 βˆ’ 𝑆1𝑃 = βˆ†π‘₯ β€’ βˆ†π‘₯ is the path difference.
Young’s Double Slit Experiment 𝐷 𝑑 𝑂 𝐡 πœƒ 𝑃 𝑦 𝑑/2 𝑑/2 ● Path difference at any general point 𝑃, βˆ†π‘₯ = 𝑆2𝑃 βˆ’ 𝑆1𝑃 βˆ†π‘₯ = 𝑦 + 𝑑 2 2 + 𝐷2 βˆ’ 𝑦 βˆ’ 𝑑 2 2 + 𝐷2 Approximation 1: 𝐷 ≫ 𝑑 Approximation 2: πœƒ is verysmall βˆ†π‘₯ β‰ˆ 𝑦𝑑 𝐷
Young’s Double Slit Experiment βˆ†π‘₯ = 𝑆2𝐴 = 𝑑 sin πœƒ πœƒ is verysmall β†’ sin πœƒ β‰ˆ tan πœƒ βˆ†π‘₯ β‰ˆ 𝑑 tan πœƒ = 𝑑𝑦 𝐷 βˆ†π‘₯ β‰ˆ 𝑦𝑑 𝐷
● For Maxima (Constructive interference): βˆ†π‘₯ = π‘›πœ† 𝑦 = π‘›πœ†π· 𝑑 Where, 𝑛 = 0, Β±1, Β±2, Β±3, … β‡’ 𝑑𝑦 𝐷 = π‘›πœ† 𝑛 = 0 correspondstothecentralmaxima. 𝑛 = Β±1 correspondtothe1𝑠𝑑 maxima. 𝑛 = Β±2 correspondtothe2𝑛𝑑 maxima. 𝑃 𝐡 𝑑 𝑂 𝑦 𝑆1 𝑆2 𝐷 βˆ†π‘₯ Condition for Constructive Interference β†’ 𝑦 = 0 β†’ 𝑦 = Β± πœ†π· 𝑑 β†’ 𝑦 = Β± 2πœ†π· 𝑑
● For Minima (Destructive interference): βˆ†π‘₯ = 𝑛 + 1 2 πœ† β‡’ 𝑑𝑦 𝐷 = 𝑛 + 1 2 πœ† 𝑦 = 𝑛 + 1 2 πœ†π· 𝑑 𝑛 = 0,βˆ’1 correspondtothe1𝑠𝑑 minima. 𝑛 = 1,βˆ’2 correspondtothe2𝑛𝑑 minima. 𝑃 𝐡 𝑑 𝑂 𝑦 𝑆1 𝑆2 𝐷 βˆ†π‘₯ Condition for Destructive Interference Where, 𝑛 = 0, Β±1, Β±2, Β±3, … β†’ 𝑦 = Β± πœ†π· 2𝑑 β†’ 𝑦 = Β± 3πœ†π· 2𝑑
In YDSE, white light is passed through the double slit and interference pattern is observed on a screen 2.5 π‘š away. The separation between the slits is 0.5 π‘šπ‘š. The first violet and red maxima are formed at distances of 2 π‘šπ‘š and 3.5 π‘šπ‘š away from the central white maxima, respectively. The wavelengths of red and violet light, respectively, are: Given: To find: Solution: 𝐷 = 2.5 π‘š, 𝑑 = 0.5 π‘šπ‘š, π‘¦π‘£π‘–π‘œπ‘™π‘’π‘‘ = 2 π‘šπ‘š, π‘¦π‘Ÿπ‘’π‘‘ = 3.5 π‘šπ‘š πœ†π‘£π‘–π‘œπ‘™π‘’π‘‘, πœ†π‘Ÿπ‘’π‘‘ Distance of first maxima from the central maxima is given by: For violet light, 𝑦 = 2 π‘šπ‘š β‡’ 2 Γ— 10βˆ’3 = 2.5 Γ— πœ†π‘£π‘–π‘œπ‘™π‘’π‘‘ 0.5 Γ— 10βˆ’3 β‡’ πœ†π‘£π‘–π‘œπ‘™π‘’π‘‘ = 2 5 Γ— 10βˆ’6 π‘š 𝑦 = π·πœ† 𝑑 For red light, 𝑦 = 3.5 π‘šπ‘š β‡’ 3.5 Γ— 10βˆ’3 = 2.5 Γ— πœ†π‘Ÿπ‘’π‘‘ 0.5 Γ— 10βˆ’3 β‡’ πœ†π‘Ÿπ‘’π‘‘= 0.7 Γ— 10βˆ’6 π‘š πœ†π‘Ÿπ‘’π‘‘ = 700 π‘›π‘š πœ†π‘£π‘–π‘œπ‘™π‘’π‘‘ = 400 π‘›π‘š
Resultant Amplitude of the Wave 𝐸1 = 𝐸01 sin π‘˜π‘₯ βˆ’ πœ”π‘‘ 𝐸2 = 𝐸02 sin π‘˜π‘₯ βˆ’ πœ”π‘‘ + 𝛿 𝐸01, 𝐸02 π‘œπ‘Ÿ 𝐴01, 𝐴02 β†’ Electric Field Amplitude. Electric field of the wave from 𝑆1: Electric field of the wave from 𝑆2: 𝐸0 2 = 𝐸01 2 + 𝐸02 2 + 2𝐸01𝐸02 cos 𝛿 𝐸𝑛𝑒𝑑 = 𝐸1 + 𝐸2 = 𝐸0 sin π‘˜π‘₯ βˆ’ πœ”π‘‘ + 𝛿 Net electric field after the interference: tan πœ€ = 𝐸02 sin 𝛿 𝐸01 + 𝐸02 cos 𝛿
Intensity of Waves 𝐼 = 2πœ‹2 𝑓2 πœŒπ‘£πΈ0 2 𝐼 ∝ 𝐸0 2 𝐼𝑛𝑒𝑑 = 𝐼1 + 𝐼2 + 2 𝐼1 𝐼2 cos 𝛿 𝐸0 2 = 𝐸01 2 + 𝐸02 2 + 2𝐸01𝐸02 cos 𝛿 𝑃 𝐡 𝑑 𝑂 𝑦 𝑆1 𝑆2 𝐷 𝐸2 𝐸1
Intensity for Identical Waves 𝐼𝑛𝑒𝑑 = 4𝐼 cos2 𝛿 2 𝐼𝑛𝑒𝑑 = 𝐼1 + 𝐼2 + 2 𝐼1 𝐼2 cos 𝛿 ● For identical slits: 𝐼1 = 𝐼2 = 𝐼 𝐼𝑛𝑒𝑑 = 𝐼 + 𝐼 + 2 𝐼 𝐼 cos 𝛿 𝐼𝑛𝑒𝑑 = 2𝐼 + 2𝐼 cos 𝛿 = 2𝐼(1 + cos 𝛿) 𝐼𝑛𝑒𝑑 = 2𝐼 Γ— 2 cos2 𝛿 2 𝑃 𝐡 𝑑 𝑂 𝑦 𝑆1 𝑆2 𝐷 𝐼2 𝐼1
● For constructive interference: (π‘€π‘Žπ‘₯π‘–π‘šπ‘Ž) Intensity of Waves ● For destructive interference: (π‘€π‘–π‘›π‘–π‘šπ‘Ž) 𝐸0 π‘šπ‘Žπ‘₯ = 𝐸01 + 𝐸02 𝐸0 π‘šπ‘–π‘› = 𝐸01 βˆ’ 𝐸02 𝐼 ∝ 𝐸0 2 β‡’ πΌπ‘šπ‘Žπ‘₯ πΌπ‘šπ‘–π‘› = 𝐸0 π‘šπ‘Žπ‘₯ 𝐸0 π‘šπ‘–π‘› 2 πΌπ‘šπ‘Žπ‘₯ πΌπ‘šπ‘–π‘› = 𝐸01 + 𝐸02 𝐸01 βˆ’ 𝐸02 2 ● When cos 𝛿 = 1, 𝛿 = 2π‘›πœ‹ (π‘€π‘Žπ‘₯π‘–π‘šπ‘Ž) ● When cos 𝛿 = βˆ’1, 𝛿 = 2𝑛 + 1 πœ‹ (π‘€π‘–π‘›π‘–π‘šπ‘Ž) 𝐼𝑛𝑒𝑑 = 𝐼1 + 𝐼2 + 2 𝐼1 𝐼2 β‡’ 𝐼1 + 𝐼2 2 𝐼𝑛𝑒𝑑 = 𝐼1 + 𝐼2 βˆ’ 2 𝐼1 𝐼2 β‡’ 𝐼1 βˆ’ 𝐼2 2 𝐼 = 𝐼1 + 𝐼2 2 , constructive interference 𝐼1 βˆ’ 𝐼2 2 , destructive interference
Two coherent sources produce waves of different intensities which interfere. After interference, the ratio of the maximum intensity to the minimum intensity is 16. The intensity of the waves are in the ratio: 𝐼1 𝐼2 = 25 9 Solution: Given: To find: πΌπ‘šπ‘Žπ‘₯ πΌπ‘šπ‘–π‘› = 16 𝐼1 𝐼2 πΌπ‘šπ‘Žπ‘₯ πΌπ‘šπ‘–π‘› = 16 β‡’ 𝐼1 + 𝐼2 𝐼1 βˆ’ 𝐼2 2 = 16 1 β‡’ 𝐼1 + 𝐼2 = 4 𝐼1 βˆ’ 4 𝐼2 β‡’ 3 𝐼1 = 5 𝐼2 β‡’ 𝐼1 𝐼2 = 5 3 Ratio of intensities: 𝐼1 𝐼2 = 5 3 2
Intensity Variation 𝑦 = π‘›πœ†π· 𝑑 𝑦 = 𝑛 + 1 2 πœ†π· 𝑑 ● Constructive Interference: ● Destructive Interference: 𝑦 1𝑠𝑑 order maxima 2nd order maxima Central maxima 1𝑠𝑑 order minima 2𝑠𝑑 order minima 𝑦 = Β± πœ†π· 𝑑 , Β± 2πœ†π· 𝑑 , Β± 3πœ†π· 𝑑 . . . . . . . 𝑦 = Β± πœ†π· 2𝑑 , Β± 3πœ†π· 2𝑑 , Β± 5πœ†π· 2𝑑 . . . . . . . 𝐷 𝑂 𝑆1 𝑆2 𝑑 𝐼0 𝐼0
Intensity Variation 𝐼 = 4𝐼0 cos2 𝛿 2 ● Intensity at any point: ● πΌπ‘šπ‘Žπ‘₯ = 4𝐼0, πΌπ‘šπ‘–π‘› = 0 𝛿 = Phase difference between the two waves from 𝑆1 and 𝑆2. 𝐼0 𝐼0 4𝐼0 𝐼 = 0 𝐼 = 0 𝐼 = 0 𝐼 = 0 𝐷 𝑂 𝑆1 𝑆2 𝑑 4𝐼0 4𝐼0 4𝐼0 4𝐼0
Shape of Fringes on Screen If we replace slits by pin holes in YDSE, then we will see a Hyperbolic Fringe pattern. Pointsourcesplacedon theperpendicularaxis tothescreencreateconcentriccircularfringes.
Fringe Width 𝐷 𝑂 𝑆1 𝑆2 𝑑 𝝀𝑫 𝒅 Fringe width is the distance between two consecutive maxima/minima. 𝛽 = 3πœ†π· 2𝑑 βˆ’ πœ†π· 2𝑑 𝛽 = πœ†π· 𝑑
Fringe Width when the setup is inside a Medium If experimental setup is dipped in liquid πœ‡ = π‘“πœ†π‘£π‘Žπ‘π‘’π‘’π‘š π‘“πœ†π‘šπ‘’π‘‘π‘–π‘’π‘š β‡’ πœ†π‘šπ‘’π‘‘π‘–π‘’π‘š = πœ†π‘£π‘Žπ‘π‘’π‘’π‘š πœ‡ 𝛽 = πœ†π· 𝑑 Fringe Width Refractive index 𝑐 is speed of light in vacuum 𝑣 is speed of light in liquid 𝛽 = πœ†π‘šπ‘’π‘‘π‘–π‘’π‘šπ· 𝑑 β‡’ 𝛽 = πœ†π‘£π‘Žπ‘π‘’π‘’π‘šπ· πœ‡π‘‘ Fringe Width inside liquid, πœ‡ = 𝑐 𝑣
Angular Fringe Width 𝑃 𝐡 𝑑 πœƒ 𝑂 𝑦 = 𝛽 𝑆1 𝑆2 𝐷 Central Maxima First Maxima tan πœƒ β‰ˆ πœƒ = 𝛽 𝐷 We know 𝛽 = πœ†π· 𝑑 After putting 𝛽 value in πœƒ = 𝛽 𝐷 πœƒ = πœ† 𝑑
Position of Maxima/Minima 𝑦 = π‘›πœ†π· πœ‡π‘‘ 𝑦 = 𝑛 + 1 2 πœ†π· πœ‡π‘‘ ● Constructive Interference: ● Destructive Interference: 𝑦 = Β± πœ†π· πœ‡π‘‘ , Β± 2πœ†π· πœ‡π‘‘ , Β± 3πœ†π· πœ‡π‘‘ . . . . . . . 𝑦 = Β± πœ†π· 2πœ‡π‘‘ , Β± 3πœ†π· 2πœ‡π‘‘ , Β± 5πœ†π· 2πœ‡π‘‘ . . . . . . . 2πœ†π· πœ‡π‘‘ 3πœ†π· 2πœ‡π‘‘
Incoherent Light Sources Incoherent Light Sources ● However, the human eye cannot the capture the rapidly changing bright and dark fringes. ● In ultra slow motion, the fringes on the screen flicker. ● So, the eyes see a continuous band of light. ● Net Intensity on the screen is the sum of intensity from two sources.
In a Young's double slit interference experiment, the fringe pattern is observed on a screen placed at a distance 𝐷 from the slits. The slits are separated by a distance 𝑑 and are illuminated by monochromatic light of wavelength πœ†. Find the distance from the central point 𝐡 where the intensity falls to half the maximum. 𝐼 = 4𝐼0 cos2 𝛿 2 β‡’ cos 𝛿 2 = 1 2 β‡’ 𝛿 = πœ‹ 2 Phase difference 𝛿 = 2πœ‹ πœ† βˆ†π‘₯ = 𝑦𝑑 𝐷 πœ† 4 = 𝑦𝑑 𝐷 β‡’ β‡’ βˆ†π‘₯ = πœ† 4 𝑦 = πœ†π· 4𝑑 πœ†π· 2𝑑 𝐷 𝑂 𝑆1 𝑆2 𝑑 πΌπ‘šπ‘Žπ‘₯ πΌπ‘šπ‘Žπ‘₯ 2 πœ†π· 4𝑑 𝐡 Path difference βˆ†π‘₯ 4𝐼0 cos2 𝛿 2 = 1 2 4𝐼0 Intensity is half the maximum, Solution: 2πœ‹ πœ† βˆ†π‘₯ = πœ‹ 2
Two coherent point sources 𝑆1 and 𝑆2 vibrating in phase emit light of wavelength πœ†. The separation between the sources is 2πœ†. Consider a line passing through 𝑆2 and perpendicular to the line 𝑆1𝑆2. What is the smallest distance from 𝑆2 where the intensity is minimum? T Solution: Given: To find: 𝑑𝑆1𝑆2 = 2πœ† π‘₯ 2πœ† 2 + π‘₯2 βˆ’ π‘₯ = 𝑛 + 1 2 πœ† When π‘₯ > 0, 16πœ† βˆ’ 2𝑛 + 1 2 πœ† > 0 So, 2𝑛 + 1 < 4 𝑛 < 3 2 ∴ 𝑛 = 1 π‘₯ = 16πœ† βˆ’ 2𝑛 + 1 2 πœ† 4 2𝑛 + 1 β‡’ 16πœ† βˆ’ 9πœ† 12 π‘₯ = 7πœ† 12 π‘₯ = 16πœ† βˆ’ 2𝑛 + 1 2 πœ† 4 2𝑛 + 1
Solution: T Optical path difference between 𝐡𝑃0 and 𝐴𝑃0 : βˆ†π‘₯ = 𝑑 sin πœƒ β‰ˆ 𝑑 tan πœƒ βˆ†π‘₯ = 𝑑 𝑑 2𝐷 = 𝑑2 2𝐷 βˆ†π‘₯ = πœ† 3 = 𝑑2 2𝐷 𝑑 = 2πœ†π· 3 Figure shows three equidistant slits being illuminated by a monochromatic parallel beam of light. Let 𝐡𝑃0 βˆ’ 𝐴𝑃0 = πœ†/3 and 𝐷 ≫ πœ†. Show that in this case 𝑑 = 2πœ†π·/3.
Path Difference between the Two Waves
Optical Path length 𝐸 = 𝐸0 sin π‘˜π‘₯0 βˆ’ πœ”π‘‘ Original wave equation: 𝐸 = 𝐸0 sin π‘˜ π‘₯0 + 𝐿 βˆ’ πœ”π‘‘ Equation of a wave when it is ahead by a length 𝐿: 𝐸 = 𝐸0 sin π‘˜π‘₯0 βˆ’ πœ”π‘‘ + π‘˜πΏ Phase difference βˆ†πœ™ π‘˜ = 2πœ‹ πœ† So, βˆ†πœ™ = 2πœ‹ πœ† 𝐿 ∴ Two points on a wave separated by a path length of 𝐿 will have a phase difference of 2πœ‹ πœ† 𝐿. 𝐸 π‘₯ 𝐿 π‘₯0 π‘₯0 + 𝐿
Optical Path length 2πœ‹ πœ†π‘Ž πΏπ‘Žπ‘–π‘Ÿ = 2πœ‹ πœ†π‘Ž πœ‡πΏπ‘šπ‘’π‘‘ Phase Difference between points 𝐴 and 𝐡 on the wave, travelling in air: βˆ†πœ™π‘šπ‘’π‘‘ = 2πœ‹ πœ†π‘šπ‘’π‘‘ πΏπ‘šπ‘’π‘‘ Phase Difference between points 𝐴 and 𝐡0 on the wave, travelling in medium: πœ†π‘šπ‘’π‘‘ = πœ†π‘Ž πœ‡ = 2πœ‹ πœ†π‘Ž πœ‡ πΏπ‘šπ‘’π‘‘ βˆ†πœ™π‘šπ‘’π‘‘ = 2πœ‹ πœ†π‘Ž πœ‡πΏπ‘šπ‘’π‘‘ βˆ†πœ™π‘Žπ‘–π‘Ÿ = βˆ†πœ™π‘šπ‘’π‘‘ If, βˆ†πœ™π‘Žπ‘–π‘Ÿ = 2πœ‹ πœ†π‘Ž πΏπ‘Žπ‘–π‘Ÿ πΏπ‘Žπ‘–π‘Ÿ = πœ‡πΏπ‘šπ‘’π‘‘ β‡’ πΏπ‘šπ‘’π‘‘ πœ‡ π‘Žπ‘–π‘Ÿ πΏπ‘Žπ‘–π‘Ÿ 𝐴 𝐡 𝐴 𝐡0 Optical Path Length in a medium is the corresponding path that light travels in vacuum to undergo the same phase difference.
𝑂𝑃𝐷 = 𝑂𝑃𝐿𝐼𝐼 βˆ’ 𝑂𝑃𝐿𝐼 = πœ‡π‘šπ‘’π‘‘ βˆ’ 1 𝐿 Optical Path Difference Optical Path Length in air: 𝑂𝑃𝐿𝐼 = 𝐴𝐡 + πœ‡π‘Žπ‘–π‘ŸπΏ + 𝐢𝐷 = 𝐴𝐡 + 𝐿 + 𝐢𝐷 Optical Path Length in medium: 𝑂𝑃𝐿𝐼𝐼 = 𝐴𝐡 + πœ‡π‘šπ‘’π‘‘πΏ + 𝐢𝐷 Optical Path Difference: Phase Difference: 𝛿 = βˆ†πœ™πΌπΌ βˆ’ βˆ†πœ™πΌ = 2πœ‹ πœ†π‘Ž πœ‡π‘šπ‘’π‘‘ βˆ’ 1 𝐿 𝛿 = 2πœ‹ πœ†π‘Ž 𝑂𝑃𝐷 𝑂𝑃𝐷 = 𝑂𝑃𝐿𝐼𝐼 βˆ’ 𝑂𝑃𝐿𝐼 = πœ‡2 βˆ’ πœ‡1 𝐿 and Phase Difference: 𝛿 = βˆ†πœ™πΌπΌ βˆ’ βˆ†πœ™πΌ βˆ†πœ™πΌπΌ = 2πœ‹ πœ†π‘Ž πœ‡2𝐿 𝛿 = 2πœ‹ πœ†π‘Ž πœ‡2 βˆ’ πœ‡1 𝐿 Phase Difference in medium 1 and 2: Optical Path Difference: βˆ†πœ™πΌ = 2πœ‹ πœ†π‘Ž πœ‡1𝐿
Thin Transparent Film in YDSE Optical Path Difference: 𝑂𝑃𝐷 = 𝐼 βˆ’ 𝐼𝐼 = πœ‡π‘‘ + 𝑆1𝐡 βˆ’ 𝑑 βˆ’ 𝑆2𝐡 Optical Path Travelled by wave 𝐼: πœ‡π‘‘ + 𝑆1𝐡 βˆ’ 𝑑 𝑆2𝐡 Optical Path Travelled by wave 𝐼𝐼: 𝑆1𝐡 = 𝑆2𝐡 𝑂𝑃𝐷 = πœ‡π‘‘ βˆ’ 𝑑 = πœ‡ βˆ’ 1 𝑑 β€’ Introduction of slab causes change in 𝑂𝑃𝐷 by πœ‡ βˆ’ 1 𝑑. β€’ Path length of the ray 𝑆1𝐡 increases when it encounters the thin film. β€’ Now the central maxima does not lie at 𝐡.
Thin Transparent Film in YDSE Optical Path Difference: βˆ†π‘₯ = 𝑆2𝑃 βˆ’ 𝑆1𝑃 βˆ†π‘₯ = 𝑦𝑑 𝐷 βˆ†π‘₯ = 𝐼𝐼 βˆ’ 𝐼 Total Optical Path Difference at point P: βˆ†π‘₯ = 𝐼𝐼 βˆ’ 𝐼 βˆ†π‘₯ = 𝑦𝑑 𝐷 βˆ’ 𝑑 πœ‡ βˆ’ 1 β€’ After inserting thin film, light ray 𝑆1𝑃 has travelled an extra path of πœ‡ βˆ’ 1 𝑑. βˆ†π‘₯ = 𝑆2𝑃 βˆ’ 𝑆1𝑃 + 𝑑 πœ‡ βˆ’ 1 β€’ After inserting thin film, light ray 𝑆2𝑃 has travelled an extra path of πœ‡ βˆ’ 1 𝑑. Total Optical Path Difference at point P: βˆ†π‘₯ = 𝐼𝐼 βˆ’ 𝐼 βˆ†π‘₯ = 𝑦𝑑 𝐷 + 𝑑 πœ‡ βˆ’ 1 βˆ†π‘₯ = 𝑆2𝑃 + 𝑑 πœ‡ βˆ’ 1 βˆ’ 𝑆1𝑃
Shift in Central Maxima 𝑃 𝐡 𝑑 πœƒ 𝑂 𝑦 π‘₯ π‘₯ + Ξ”π‘₯ 𝐷 𝑑, πœ‡ πœ‡ βˆ’ 1 𝑑𝐷 𝑑 At central maxima, 𝑆1 𝑆2 𝐼 𝐼𝐼 Screen 𝑂𝑃𝐷 = 0 𝑦𝑑 𝐷 βˆ’ πœ‡ βˆ’ 1 𝑑 = 0 𝑦 = πœ‡ βˆ’ 1 𝑑𝐷 𝑑 Shift in central maxima =
Number of Fringes Shifted πœ‡ βˆ’ 1 𝑑𝐷 𝑑 Shift in central maxima: Number of Fringes shifted 𝑛 = 𝑦 𝛽 = πœ‡ βˆ’ 1 𝑑𝐷 𝑑 Γ— 𝑑 πœ†π· 𝑛 = πœ‡ βˆ’ 1 𝑑 πœ† = π‘ β„Žπ‘–π‘“π‘‘ 𝑖𝑛 π‘π‘’π‘›π‘‘π‘Ÿπ‘Žπ‘™ π‘šπ‘Žπ‘₯π‘–π‘šπ‘Ž π‘“π‘Ÿπ‘–π‘›π‘”π‘’ π‘€π‘–π‘‘π‘‘β„Ž 𝑆 = πœ‡ βˆ’ 1 𝑑𝐷 𝑑 Shift in central maxima: Number of fringes shifted: 𝑛 = πœ‡ βˆ’ 1 𝑑 πœ†
T The Young’s double slit experiment is done in a medium of refractive index 4/3. A light of 600 π‘›π‘š wavelength is falling on the slits having 0.45 π‘šπ‘š separation. The lower slit 𝑆2 is covered by a thin glass sheet of thickness 10.2 πœ‡π‘š and refractive index 1.5. The interference pattern is observed on a screen placed 1.5 π‘š from the slits. Find the location of the central maximum on the 𝑦 βˆ’axis. 10.2π‘šπ‘š 7.5 π‘šπ‘š 1.25 π‘šπ‘š 4.25 π‘šπ‘š B A D C 𝑃0 𝑂 πœ‡ = 4/3 𝑆1 𝑆2 Screen 𝐼 𝐼𝐼
Given: Solution: To find: πœ‡π‘šπ‘’π‘‘ = 4 3 , 𝑑 = 0.45 π‘šπ‘š, 𝐷 = 1.5 π‘š, πœ† = 600 π‘›π‘š, 𝑑 = 10.2 πœ‡π‘š, πœ‡π‘‘ = 1.5 Position of central maxima 𝑂𝑃𝐷 = 𝐼 βˆ’ 𝐼𝐼 𝑂𝑃𝐷 = 𝑆1𝑃0 πœ‡π‘šπ‘’π‘‘ βˆ’ πœ‡π‘”π‘‘ + 𝑆2𝑃0 πœ‡π‘šπ‘’π‘‘ βˆ’ πœ‡π‘šπ‘’π‘‘π‘‘ 𝑂𝑃𝐷 = 𝑆1𝑃0 βˆ’ 𝑆2𝑃0 πœ‡π‘šπ‘’π‘‘ βˆ’ 𝑑 πœ‡π‘” βˆ’ πœ‡π‘šπ‘’π‘‘ 𝑂𝑃𝐷 = 𝑦𝑑 𝐷 πœ‡π‘šπ‘’π‘‘ βˆ’ 𝑑 πœ‡π‘” βˆ’ πœ‡π‘šπ‘’π‘‘ At central maxima 𝑂𝑃𝐷 = 0 𝑦𝑑 𝐷 πœ‡π‘šπ‘’π‘‘ βˆ’ 𝑑 πœ‡π‘” βˆ’ πœ‡π‘šπ‘’π‘‘ = 0 𝑦 = 𝑑𝐷 π‘‘πœ‡π‘šπ‘’π‘‘ πœ‡π‘” βˆ’ πœ‡π‘šπ‘’π‘‘ Substituting all the values we get, 𝑦 = 4.25 π‘šπ‘š 𝑂 πœ‡ = 4/3 𝑆1 𝑆2 Screen 𝑃0 𝐼 𝐼𝐼 𝑦
Given: Solution: To find: πœ† = 5000 ሢ 𝐴, πœ‡ = 1.5, 𝑑 𝑦 = πœ‡ βˆ’ 1 𝑑𝐷 𝑑 𝑦5𝐡 = 5π·πœ† 𝑑 𝑦𝑛𝐡 = π‘›π·πœ† 𝑑 Shift in central maxima: πœ‡ βˆ’ 1 𝑑𝐷 𝑑 = 5π·πœ† 𝑑 𝑑 = 5πœ† πœ‡ βˆ’ 1 = 5 Γ— 5000 Γ— 10βˆ’10 1.5 βˆ’ 1 𝑑 = 5 πœ‡π‘š T In YDSE, find the thickness of a glass slab (πœ‡ = 1.5) which should be kept in front of upper slit 𝑆1 so that the central maxima is formed at a place where 5π‘‘β„Ž bright fringe was lying earlier (before inserting slab). πœ† = 5000 ሢ 𝐴
Given: Solution: To find: Wavelength = πœ†, 𝑑1 = 𝑑2 = 𝑑 =? Path difference due to both the slabs at 𝑃0: For 𝑑 to be minimum, put 𝑛 = 0, β‡’ πœ‡1 βˆ’ πœ‡2 𝑑 = 𝑛 + 1 2 πœ† Ξ”π‘₯ = 𝑛 + 1 2 πœ† For minima at 𝑃0, we know, πœ‡1 βˆ’ 1 𝑑 βˆ’ πœ‡2 βˆ’ 1 𝑑 = πœ‡1 βˆ’ πœ‡2 𝑑 Refractive index = πœ‡1 & πœ‡2 Screen 𝐷 𝑃0 𝑑 𝑂 𝑆1 𝑆2 𝑑 πœ‡1 πœ‡2 T 𝑑 = πœ† 2 πœ‡1 βˆ’ πœ‡2 Two transparent slabs, having equal thickness but different refractive indices πœ‡1 and πœ‡2 πœ‡1 > πœ‡2 , are pasted side by side to form a composite slab. This slab is placed just after the double slits in a Young’s experiment so that the light from one slit goes through one material and light through other slit goes through other material. What should be the minimum thickness of the slab so that there is a minimum at point 𝑃0 which is equidistant from the slits?
Phase change Yes No Reflected from Denser Medium Rarer Medium 𝑖 π‘Ÿ Air Air πœ‡ No phase change No phase change πœ‹ Phase change in Reflection & Refraction
A narrow slit 𝑆 transmitting light of wavelength πœ† is placed at a distance 𝑑 above a large plane mirror. The light coming directly from the slit and that coming after reflection interfere at a screen placed at a distance 𝐷 from the slit. (π‘Ž) What will be the intensity at a point just above 𝑂. Given: To find: Solution: Wavelength = πœ†, Distance of screen = 𝐷, 𝑆𝑃 β‰ˆ 𝑆𝑄𝑃 β†’ path difference = 0 (when 𝑃 is just above 𝑂) Screen 𝑂 𝐷 𝑆 𝑑 𝑄 𝑃 Ξ΄ = πœ‹ (due to reflection of light from a denser medium) Distance of source from the mirror = 𝑑 Intensity just above 𝑂 T 𝐼𝑛𝑒𝑑 = 0 If phase difference is odd integral multiple of πœ‹, destructive interference will take place.
A narrow slit 𝑆 transmitting light of wavelength πœ† is placed at a distance 𝑑 above a large plane mirror. The light coming directly from the slit and that coming after reflection interfere at a screen placed at a distance 𝐷 from the slit. Given: To find: Solution: T (𝑏) At what distance from 𝑂 does the first maximum occur? Screen 𝑂 𝐷 𝑆1 𝑑 𝑄 𝑃 𝑆2 𝑑 πœƒ 𝑦 Wavelength = πœ†, 𝑦1𝑠𝑑 π‘šπ‘Žπ‘₯π‘–π‘šπ‘Ž Distance of screen = 𝐷, Distance of source from the mirror = 𝑑 βˆ†π‘₯ = 2𝑑 sin πœƒ = 2𝑑 tan πœƒ = 2𝑑𝑦 𝐷 βˆ†π‘₯𝑛𝑒𝑑 = 2𝑑𝑦 𝐷 + πœ† 2 = π‘›πœ† = 1 πœ† β‡’ 2𝑑𝑦 𝐷 = πœ† 2 𝑦 = πœ†π· 4𝑑
Given: To find: Solution: πœ† = 700 π‘›π‘š, 𝐷 = 1 π‘š, 𝛽 = 2 Γ— πœ†π· 4𝑑 = πœ†π· 2𝑑 𝛽 = 700 Γ— 10βˆ’9 Γ— 1 2 Γ— 10βˆ’3 = 3.5 Γ— 10βˆ’4 π‘š 𝑑 = 1 π‘šπ‘š Fringe width = 𝛽 T A narrow slit 𝑆 transmitting light of wavelength πœ† is placed at a distance 𝑑 above a large plane mirror. The light coming directly from the slit and that coming after reflection interfere at a screen placed at a distance 𝐷 from the slit. (𝑐) If 𝑑 = 1 π‘šπ‘š, 𝐷 = 1 π‘š and πœ† = 700 π‘›π‘š, then find the fringe width. 𝑂 𝐷 𝑆1 𝑑 𝑄 𝑃 𝑆2 𝑑 πœƒ πœ†π· 4𝑑 πœ†π· 4𝑑 𝛽 = 0.35 π‘šπ‘š
A narrow slit 𝑆 transmitting light of wavelength πœ† is placed at a distance 𝑑 above a large plane mirror. The light coming directly from the slit and that coming after reflection interfere at a screen placed at a distance 𝐷 from the slit. (𝑑) If the mirror reflects only 64 % of the light energy falling on it, what will be the ratio of the maximum to the minimum intensity in the interference pattern observed on the screen? Given: To find: Solution: πΌπ‘šπ‘Žπ‘₯ = 𝐼1 + 𝐼2 2 , πΌπ‘šπ‘Žπ‘₯ πΌπ‘šπ‘–π‘› T πΌπ‘šπ‘Žπ‘₯ πΌπ‘šπ‘–π‘› = 81 𝑂 𝐷 𝑆1 𝑄 𝑃 𝑆2 πœƒ 𝐼1 = 𝐼0 𝐼2 = 0.64 𝐼0 Wavelength = πœ†, Distance of screen = 𝐷, Distance of source from the mirror = 𝑑 𝐼2 = 0.64 𝐼0 πΌπ‘šπ‘–π‘› = 𝐼1 βˆ’ 𝐼2 2 πΌπ‘šπ‘Žπ‘₯ πΌπ‘šπ‘–π‘› = 𝐼1 + 𝐼2 2 𝐼1 βˆ’ 𝐼2 2 β‡’ 𝐼0 + 0.64 𝐼0 2 𝐼0 βˆ’ 0.64 𝐼0 2 = 1 + 0.8 2 1 βˆ’ 0.8 2 = 1.8 2 0.2 2
Condition for Thin Film Interference 𝑑 𝑖 π‘Ÿ air air πœ‡ πœ‹ β€’ The film thickness should be comparable to the wavelength of light 𝑑 β‰ˆ πœ† . β€’ The incident light should be white (non - monochromatic). β€’ The angle of incidence should be small 𝑖 β‰ˆ 0.
Thin Film Interference β€’ Inside the film, when a particular colour’s path difference is even integral multiple of the wavelength, it undergoes constructive interference, so these colours appear bright. β€’ Interference of light wave being reflected off two surfaces that are at a distance comparable to its wavelength is known as thin film ”Thin film interference”. β€’ When a particular colour’s wavelength is odd integral multiple of the path difference inside the film, it undergoes destructive interference, so these colours don’t appear at all. 𝑖 π‘Ÿ Air Air πœ‡ No phase change No phase change πœ‹
βˆ†π‘₯ = 2πœ‡π‘‘ Interference due to Thin Film from Transmitted Light angles are very small 𝑑 𝑖 π‘Ÿ air air πœ‡ Path difference of 1𝑠𝑑 and 2𝑛𝑑 transmitted light wave π‘Ÿ 𝑑 𝑑 cos π‘Ÿ = 2 𝑑 cos π‘Ÿ πœ‡ When angle is very small, π‘Ÿ β‰ˆ 0 cos π‘Ÿ β‰ˆ 1 Path difference π‘Ÿ 1𝑠𝑑 2𝑛𝑑
For constructive interference: Or For destructive interference: Or 2πœ‡π‘‘ = π‘›πœ† 2πœ‡π‘‘ = 𝑛 + 1 2 πœ† 𝛿 = 2π‘›πœ‹ 𝛿 = 2𝑛 + 1 πœ‹ angles are very small 𝑑 𝑖 π‘Ÿ air air πœ‡ Interference due to Thin Film from Transmitted Light
βˆ†π‘₯ = 2πœ‡π‘‘ βˆ’ πœ† 2 Path difference of 1𝑠𝑑 and 2𝑛𝑑 reflected light wave π‘Ÿ 𝑑 𝑑 cos π‘Ÿ = 2 𝑑 cos π‘Ÿ πœ‡ β‰ˆ 2πœ‡π‘‘ Total path difference 𝑑 𝑖 π‘Ÿ air air πœ‡ angles are very small 1𝑠𝑑 2𝑛𝑑 Interference due to Thin Film from Reflected Light Because phase change of πœ‹ after reflection of 1𝑠𝑑 , path difference = βˆ’ πœ† 2
For constructive interference: For destructive interference: 2πœ‡π‘‘ = π‘›πœ† 2πœ‡π‘‘ = 𝑛 + 1 2 πœ† 𝑑 𝑖 π‘Ÿ air air πœ‡ πœ‹ angles are very small Interference due to Thin Film from Reflected Light 2πœ‡π‘‘ βˆ’ πœ† 2 = π‘›πœ† 2πœ‡π‘‘ βˆ’ πœ† 2 = 𝑛 + 1 2 πœ† β‡’ 2πœ‡π‘‘ = 𝑛 + 1 πœ† 𝑛 is an integer.
Interference due to Thin Film For constructive interference: β€’ Colours will be strongly reflected/transmitted. Destructive interference: β€’ Colours will be poorly reflected/transmitted. This gives coloured appearance of the film.
A soap film of refractive index 1.33 is illuminated by the light of wavelength 400 π‘›π‘š at an angle of 45Β°. If there is complete destructive interference then, find the thickness of the film. Given: To find: Solution: T πœ‡ = 1.33, 𝑖 = 45Β°, πœ† = 400 π‘›π‘š 𝑑 πœ‡ = sin 𝑖 sin π‘Ÿ β‡’ 1.33 = sin 45 sin π‘Ÿ sin π‘Ÿ = 3 4 2 cos π‘Ÿ = 1 βˆ’ sin2 π‘Ÿ For Destructive interference: 2πœ‡π‘‘ cos π‘Ÿ = π‘›πœ† β‡’ 2 1.33 𝑑 0.85 = 1 400 Γ— 10βˆ’9 β‡’ 𝑑 = 1.27 Γ— 10βˆ’7 π‘š β‡’ cos π‘Ÿ = 0.85 45Β° soap film
Diffraction β€’ Bending of a wave or its deviation from its original direction of propagation while passing through a small obstruction is known as diffraction. Original direction β€’ Diffraction is explained by wave nature of light.​ β€’ Condition for bending: size of obstacle β‰ˆ πœ† β€’ Every point on the wavefront makes a secondary wave according to Huygens Principle.
β€’ Source and screen are at infinite distance from diffraction element. β€’ Source and screen are at finite distance from diffraction element. β€’ High intense interference pattern is observed on screen. β€’ Diffraction was first demonstrated by this experiment. 𝑆𝑂 𝑆𝑂 Diffraction of Light Waves Fraunhofer Diffraction Fresnel Diffraction
𝑃𝑂 𝑏 𝑃 πœƒ Screen 𝑏 2 𝑏 2 sin πœƒ πœƒ β€’ Path difference between the waves, βˆ†x = 𝑏 2 sin πœƒ Fraunhofer Diffraction – Path Difference
𝑃𝑂 𝑏 𝑃 πœƒ Screen β€’ The condition for 1𝑠𝑑 dark fringe formed at point 𝑃: Fraunhofer Diffraction – first Dark Fringe 𝑏 2 sin πœƒ = πœ† 2 𝑏 sin πœƒ = πœ† β€’ The condition for π‘›π‘‘β„Ž dark fringe: 𝑏 sin πœƒ = π‘›πœ† β€’ Central Maxima is at 𝑃0, where πœƒ = 0
𝑃𝑂 𝑏 𝑃 πœƒ Screen β€’ Electric Field Amplitude at 𝑃: Fraunhofer Diffraction – Intensity at general point 𝐸′ = 𝐸0 sin 𝛽 𝛽 β€’ When πœƒ = 0, 𝛽 = 0, 𝐸′ = 𝐸0 𝐼 = 𝐼𝑂 sin2 𝛽 𝛽2 Where 𝐼0 is the intensity at Central Maxima Where 𝛽 = 2πœ‹ πœ† 𝑏 2 sin πœƒ 𝛽 = phase difference β€’ Intensity ∝ 𝐸2 β€’ Intensity at a general point:
Fraunhofer Diffraction – Graph of Intensity 𝐼 = 𝐼𝑂 sin2 𝛽 𝛽2 𝛽 = πœ‹π‘ sin πœƒ πœ† πΌπ‘œ βˆ’ 2πœ† 𝑏 sin πœƒ 𝐼 0 βˆ’ πœ† 𝑏 πœ† 𝑏 2πœ† 𝑏 𝐼𝑂 22 𝐼𝑂 62.5 β€’ sin πœƒ = 0 ; Central Maxima β€’ sin πœƒ = ; Minima π‘›πœ† 𝑏 β€’ sin πœƒ = Β± πœ† 𝑏 , Β± 2πœ† 𝑏 , Β± 3πœ† 𝑏 … … … … β€’ Maximum intensity is distributed between βˆ’ πœ† 𝑏 and πœ† 𝑏 . β€’ If 𝑏 is large, πœ† 𝑏 β†’ 0, then Single fringe is observed and no diffraction is seen.
Fraunhofer Diffraction – Graph of Intensity 𝑏 β€’ Fraunhofer Diffraction ⟢ Intensity decreases away from the centre. β€’ 𝐼0 β†’ Intensity at the central bright fringe β€’ YDSE ⟢ Intensity is same for all fringes. β€’ 𝐼0 β†’ Intensity from a single slit. 𝐼 = 𝐼𝑂 sin2 𝛽 𝛽2 𝐼 = 4𝐼𝑂 cos2 πœ™ 2
Diffraction Difference between Interference and Diffraction Interference It is the phenomenon of superposition of two waves coming from two different coherent sources. It is the phenomenon of superposition of two waves coming from two different parts of the same wavefront. In interference pattern, all bright fringes are equally bright and equally spaced. All bright fringes are not equally bright and equally wide. Brightness and width decreases with the angle of diffraction. All dark fringes are perfectly dark. All dark fringes are perfectly dark, but their contrast with bright fringes and width decreases with angle of diffraction. In interference, bright and dark fringes are large in number for a given field of view. In diffraction, bright and dark fringes are fewer for a given field of view.
A beam of light of wavelength 600 π‘›π‘š from a distant source falls on a single slit 1 π‘šπ‘š wide and a resulting diffraction pattern is observed on a screen 2 π‘š away. The distance between the first dark fringes on either side of central bright fringe is 1π‘šπ‘š 𝛼 πœƒ 𝐷 = 2 π‘š πœƒ sin πœƒ β‰ˆ πœƒ Solution: πœƒ = πœ† 𝑏 𝛼 = 2πœƒ = 2πœ† 𝑏 𝛼 = 𝑦 𝐷 = 2πœ† 𝑏 ⟹ 𝑦 = 2πœ†π· 𝑏 𝑦 = 2 Γ— 600 Γ— 10βˆ’9 Γ— 2 10βˆ’3 = 24 Γ— 10βˆ’4 π‘š 𝑦 = 2.4 π‘šπ‘š First maxima is formed at πœ† 𝑏 distance away on both side of central maxima. 𝑦 (πœƒ is small)
𝐴 𝐡 𝑁 3πœ† 2 𝑏 𝑏 sin πœƒ = 3πœ† 2 πœƒ1𝐡 = 3πœ† 2𝑏 𝑦1𝐡 = 3π·πœ† 2𝑏 β€’ Condition for 1𝑠𝑑 secondary maxima (bright): β€’ The angle of diffraction (πœƒ1B) for 1st maximum (bright) is: β€’ The position of 1st maximum (bright) from the centre of the screen is: Single Slit Diffraction – πŸπ’”π’• Secondary Maxima
Screen First secondary maxima Second secondary maxima Central maxima First secondary maxima Second secondary maxima Where, 𝑛 = 1, 2, 3, 4 … … … β€’ The angle of diffraction πœƒπ‘› for π‘›π‘‘β„Ž maxima (bright) is: β€’ The position of π‘›π‘‘β„Ž maxima (bright) from the centre of the screen is: πœƒπ‘›π΅ = 2𝑛 + 1 πœ† 2𝑏 𝑦𝑛𝐡 = 2𝑛 + 1 π·πœ† 2𝑏 Single Slit Diffraction – 𝒏𝒕𝒉 Secondary Maxima
β€’ Angular position for minima + πœ† 𝑏 βˆ’ πœ† 𝑏 + 2πœ† 𝑏 βˆ’ 2πœ† 𝑏 β€’ Angular position for maxima sin πœƒ = Β± πœ† 𝑏 , Β± 2πœ† 𝑏 , … … , Β± π‘›πœ† 𝑏 sin πœƒ = Β± 3πœ† 2𝑏 , Β± 5πœ† 2𝑏 , … … , Β± (2𝑛 + 1)πœ† 2𝑏 Single Slit Diffraction + 3πœ† 2𝑏 βˆ’ 3πœ† 2𝑏 + 5πœ† 2𝑏 βˆ’ 5πœ† 2𝑏 0
Given: To find: Solution: Angular position of first minima is given by, sin πœƒ1𝐷 = πœ† π‘Ž ⟹ 1 2 = πœ† π‘Ž Angular position of first secondary maxima is given by, sin πœƒ1𝐡 = 3πœ† 2π‘Ž ⟹ sin πœƒ1𝐡 = 3 4 πœƒ1𝐡 = sinβˆ’1 3 4 T πœƒ1𝐷 = 30Β°; πœ† = 5000 ሢ 𝐴 Angular position of first secondary maximum (πœƒ1𝐡) In a diffraction pattern due to a single slit of width π‘Ž, the first minima is observed at an angle 30Β° when the light of wavelength 5000 ሢ 𝐴 is incident on the slit. The first secondary maxima is observed at an angle of:
β€’ When a monochromatic light is incident on the hole we see concentric circular bright and dark spots on the screen. β€’ The size of hole should be comparable to the wavelength of incident light. β€’ Central bright spot contains the most energy. β€’ Brightness of the rings decreases as we move away from the centre. β€’ For the first dark ring, sin πœƒ = 1.22πœ† 𝑏 Fraunhofer Diffraction for Hole 𝑏 𝐷 πœƒ 𝑅 𝑅 = 1.22πœ†π· 𝑏
A convex lens of diameter 8.0 π‘π‘š is used to focus a parallel beam of light of wavelength 6200 Γ…. If the light be focused at a distance of 20 π‘π‘š from the lens, what would be the radius of the central bright spot? 6200 Γ… 𝑏 = 8 π‘π‘š πœƒ 20 π‘π‘š Given: To Find: Solution: 𝐷 = 20 π‘π‘š, 𝑏 = 8 π‘π‘š and πœ† = 6200 Γ… 𝑅 Radius of central bright spot, 𝑅 = 1.22πœ†π· 𝑏 β‡’ 𝑅 = 1.22 Γ— 6200 Γ— 10βˆ’10 Γ— 20 Γ— 10βˆ’2 8 Γ— 10βˆ’2 𝑅 = 1.89 Γ— 10βˆ’6 π‘š 𝑅
Binary Star
Limit of Resolution Just resolved: Well resolved: Unresolved: Diffraction discs from both sources overlap. 𝑆1 𝑆2 𝑆1 𝑆2 𝑆1 𝑆2 The periphery of the diffraction disc of one object touches the centre of the diffraction disc of the other object. The diffraction discs formed by two objects are well separated from each other.
Rayleigh criterion The Rayleigh criterion specifies the minimum separation between two light sources that may be resolved into distinct objects. Unresolved Just resolved Well resolved (Rayleigh criterion)
Limit of Resolution of a Telescope β€’ Clear image is formed when the diffraction discs from two sources is just resolved. π‘†π‘‘π‘Žπ‘Ÿ1 π‘†π‘‘π‘Žπ‘Ÿ2 Just resolved 𝑅 βˆ†πœƒ Telescope lens β€’ Distance between diffraction disc > 𝑅 β‡’ Well resolved β€’ Distance between diffraction disc < 𝑅 β‡’ Unresolved
Resolving Power of Telescope β€’ Angular limit of resolution of telescope: βˆ†πœƒ = 1.22πœ† 𝑏 βˆ†πœƒ = Limit of resolution β€’ Angle subtended by the first dark fringe > βˆ†πœƒ β‡’ Well resolved β€’ Angle subtended by the first dark fringe < βˆ†πœƒ β‡’ Unresolved 𝑏 𝑓 𝑅 βˆ†πœƒ
Radius of the Central Bright Spot β€’ Radius of central bright region is: β€’ Resolving power of a telescope: 𝑅 = 1.22πœ†π‘“ 𝑏 𝑅. 𝑃. = 𝑏 1.22πœ† 𝑅 = π‘“βˆ†πœƒ 𝑅. 𝑃. = 1 βˆ†πœƒ β€’ Bigger lens β‡’ larger 𝑏 β‡’ smaller Limit of Resolution βˆ†πœƒ β‡’ higher Resolving Power 𝑅. 𝑃. 𝑏 𝑓 𝑅 βˆ†πœƒ
Disadvantages of using Lens 1. Difficult and expensive to build large lenses. 2.Providing mechanical support to large lenses require large and complex machinery. 3. Chromatic aberration ( light rays passing through a lens focus at different points, depending on their wavelength). The largest lens objective in use has diameter of 40 inch (~1.02 π‘š). It is at the Yerkes Observatory in Wisconsin, USA. Fact: Chromatic aberration
Advantages of using Mirror as the Objective in Telescope 1. No chromatic aberration 2. Parabolic mirror used to counter spherical aberration. 3. Large mirrors can be supported from the back. The viewer sits near the focal point of the mirror, in a small cage. Fact: Eyepiece Objective mirror
Given: To find: Solution: Limit of resolution of telescope is given by: βˆ†πœƒ = 1.22 Γ— 500 Γ— 10βˆ’9 200 Γ— 10βˆ’2 T 𝑏 = 200 π‘π‘š, πœ† = 500 π‘›π‘š Limit of resolution of telescope 305 Γ— 10–9 π‘Ÿπ‘Žπ‘‘π‘–π‘Žπ‘› βˆ†πœƒ = 1.22πœ† 𝑏 Calculate the limit of resolution of a telescope objective having a diameter of 200 π‘π‘š, if it has to detect light of wavelength 500 π‘›π‘š coming from a star. JEE Main 2019
Limit of Resolution of a Microscope Angular limit of resolution of microscope: sin πœƒ β‰ˆ πœƒ = 1.22πœ† 𝑏 𝑅 = π‘£πœƒ = 𝑣 Γ— 1.22πœ† 𝑏 π‘š = 𝑅 𝑑 β‡’ 𝑑 = 𝑅 π‘š Magnification of convex lens: 𝑑 = 𝑣 Γ— 1.22πœ† π‘π‘š β€’ Clear images can be seen in the microscope if the diffraction discs are just resolved. Lens Formula: 1 𝑣 βˆ’ 1 𝑒 = 1 𝑓 1 βˆ’ 𝑣 𝑒 = 𝑣 𝑓 π‘š = 1 βˆ’ 𝑣 𝑓 β‡’ 1 βˆ’ π‘š = 𝑣 𝑓 π‘š = 𝑣 𝑒 π‘š β‰ˆ βˆ’ 𝑣 𝑓 𝑣 ≫ 𝑓 β‡’ 𝑣 𝑓 ≫ 1 𝑑 = 𝑣 Γ— 1.22πœ† π‘π‘š We know that: 𝑑 = 1.22πœ† π‘π‘š Γ— βˆ’π‘šπ‘“ = 1.22πœ†π‘“ 𝑏 tan 𝛽 = 𝑏 2𝑓 β‡’ 𝑏 = 2𝑓 tan 𝛽 𝛽 is small, 𝑏 = 2𝑓 sin 𝛽 π‘‘π‘šπ‘–π‘› = 1.22πœ†π‘“ 𝑏 = 1.22πœ†π‘“ 2𝑓 sin 𝛽 π‘‘π‘šπ‘–π‘› = 1.22πœ† 2 sin 𝛽
Microscope immersed in Oil When the setup is immersed in oil, πœ†π‘šπ‘’π‘‘ changes to πœ† πœ‡ π‘‘π‘šπ‘–π‘› = 1.22πœ†π‘šπ‘’π‘‘ 2 sin 𝛽 ⟢ π‘‘π‘šπ‘–π‘› = 1.22πœ† 2πœ‡ sin 𝛽 Note: The product πœ‡sin𝛽 is called the numerical aperture and is sometimesmarked on the objective. Resolving power of a microscope: β€’ Resolving power of the microscope increases when it is immersed in a medium. β‡’ 𝑅. 𝑃. = 2πœ‡ sin 𝛽 1.22πœ† 𝑅. 𝑃. = 1 π‘‘π‘šπ‘–π‘›
Validity of Ray Optics β€’ Consider a single slit of width β€˜π‘Žβ€™.Diffraction pattern will be observed. β€’ There will be central maxima due to diffraction. β€’ Angular size of central maximum, πœƒ = πœ† π‘Ž If 𝑦 β‰ˆ π‘Ž β€’ The width of diffracted beam after it has travelled by 𝑧, 𝑦 = 𝑧 Γ— πœƒ = πœ†π‘§ π‘Ž 𝑧𝐹 is Fresnel distance. β€’ If 𝑧 < 𝑧𝐹, the ray optics is valid. β€’ If 𝑧 > 𝑧𝐹, spreading due to diffraction dominates. 𝑧 β‰ˆ π‘Ž2 πœ† = 𝑧𝐹
Given: 𝑧𝐹 = 2 Γ— 10βˆ’3 2 600 Γ— 10βˆ’9 = 6.7 π‘š T π‘Ž = 2 π‘šπ‘š; πœ† = 600 π‘›π‘š Solution: Ray optics is a good approximation up to Fresnel distance only. 𝑧 β‰ˆ π‘Ž2 πœ† = 𝑧𝐹 β€’ If 𝑑 < 6.7 π‘š ⟢ Ray optics holds. β€’ If 𝑑 > 6.7 π‘š ⟢ Spreading due to diffraction dominates. For what distance is ray optics a good approximation when a plane light wave is incident on a circular aperture of width 2 π‘šπ‘š having wavelength 600 π‘›π‘š?
When a parallel beam of light passes through a medium, a part of it appears in directions other than the incident direction. This phenomenon is called scattering of light. Aparticle Scattering β€’ Light is an EM wave, it oscillates the charged particles in a medium because of its oscillating electric field. β€’ Oscillating charged particles emit EM waves. β€’ If the oscillating electric field of incident light has frequency 𝑓, the frequency of scattered wave will also have frequency 𝑓. Incidentlight
Rayleigh’s Law of Scattering β€’ When size of particles < πœ† β€’ Intensity of scattering depends on Intensity of scattered wave ∝ 1 πœ†4 1. Wavelength of light 2. Size of particles causing scattering 𝑉 𝐼 𝐡 π‘Œ 𝑂 𝑅 𝐺 πœ† βˆ’ π‘šπ‘–π‘›π‘–π‘šπ‘’π‘š πœ† βˆ’ π‘šπ‘Žπ‘₯π‘–π‘šπ‘’π‘š Scatters most Scatters least πœ† increases and Intensity of scattering decreases
Unpolarized Light ● The light having electric field oscillations in all directions in the plane perpendicular to the direction of propagation. Note: Light wave is coming out of the screen, and arrows show direction of oscillation of electric field. ● Examples of source of unpolarized light: Candle, Bulb, Sun etc.
Plane Polarized Light Plane Polarized light – When electric field at a point always remains parallel to a fixed direction as time passes. Plane of polarization – Plane containing electric field and direction of propagation. π‘₯ βˆ’ π‘Žπ‘₯𝑖𝑠 β†’ Direction of propagation. 𝑦 βˆ’ π‘Žπ‘₯𝑖𝑠 β†’ Oscillation of electric field. π‘₯𝑦 π‘π‘™π‘Žπ‘›π‘’ β†’ Plane of polarization. When the polarizer is placed in the path of unpolarized light, the direction of oscillation of the electric field becomes parallel to the transmission axis. Note: If any unpolarized light of intensity 𝐼0 is incident on a polarizer, we get a polarized light of intensity . 𝐼0 2
Law of Malus ● 𝐼 = Intensity of transmitted light ● 𝐼0 = Intensity of incident light Intensity of the wave after crossing the polarizer: Electric field amplitude of the wave after crossing the polarizer: 𝐸 = 𝐸0 cos πœƒ 𝐼 ∝ 𝐸2 𝐼 = 𝐼0 cos2 πœƒ πœƒ 𝐸0 𝐸0 cos πœƒ 𝐼0 𝐼 Polarized light Unpolarized light 2𝐼0
Law of Malus 𝐼 = 𝐼0 cos2 0Β° = 𝐼0 ● Case 1 β†’ when πœƒ = 0Β° : ● Case 2 β†’ when πœƒ = 90Β° : 𝐼 = 𝐼0 cos2 90Β° = 0 𝐼0 𝐼0 𝐼0
T Two β€˜crossed’ polaroids 𝐴 and 𝐡 are placed in the path of a light beam. In between these, a third polaroid 𝐢 is placed whose polarization axis makes an angle πœƒ with the polarization axis of the polaroid 𝐴. If the intensity of light emerging from the polaroid 𝐴 is πΌπ‘œ, then the intensity of light emerging from polaroid 𝐡 will be Solution: Intensity of light emerging from polaroid 𝐢: Intensity of light emerging from polaroid 𝐡: 𝐼𝐡 = 𝐼𝐢 cos2 90Β° βˆ’ πœƒ 𝐼𝐡 = 𝐼0 cos2 πœƒ . cos2 90Β° βˆ’ πœƒ 𝐼𝐡 = 𝐼0 cos2 πœƒ . sin2 πœƒ {Applying the law of Malus.} = 𝐼0 4 2 sin πœƒ cos πœƒ 2 𝐼𝐢 = 𝐼0 cos2 πœƒ 𝐼𝐡 = 𝐼0 4 sin2 (2πœƒ)
Unpolarized light with amplitude 𝐴0 passes through two polarizers. The first one has an angle of 30Β° clockwise to vertical and second one has an angle of 15Β° counter-clockwise to the vertical. What is the amplitude of the light emitted from the second polarizer? Given: To find: 𝐴 = 𝐴0 𝐴2 Solution: We know that: 𝐼 ∝ 𝐴2 . ∴ 𝐴 ∝ 𝐼 β‡’ 𝐴1 𝐴2 = 𝐼1 𝐼2 𝐴1 𝐴0 = 𝐼1 𝐼0 = 𝐼0 2𝐼0 = 1 2 𝐴1 = 𝐴0 2 𝐼1 = 𝐼0 2 T 𝐴1 = 𝐴0 2 𝐼2 = 𝐼1cos2 πœƒ 𝐼2 = 𝐼0 2 cos2 45Β° = 𝐼0 2 1 2 2 = 𝐼0 4 𝐴2 𝐴0 = 𝐼2 𝐼0 = 𝐼0 4𝐼0 = 1 4 𝐴2 = 𝐴0 2 𝐴1 𝐴2 = 𝐼1 𝐼2 Again, πœƒ = 30Β° βˆ’ βˆ’15Β° = 45Β°
Medium(πœ‡) 𝑖 Oscillations βŠ₯ to the incidence plane Oscillations in the incidence plane Polarization by Reflection ● The reflected light has more vibrations perpendicular to plane of incidence. ● The refracted light has more vibration parallel to the plane of incidence. ● The percentage of polarization in reflected light changes as we change the angle of incidence 𝑖.
● For a particular angle of incidence 𝑖𝐡 , the reflected light becomes completely plane polarized. ● The required condition for this purpose is: 𝑖𝐡 + π‘Ÿ = 90Β° ● Brewster’s Law tan 𝑖𝐡 = πœ‡ 𝑖𝐡 = Brewster angle/polarising angle Brewster’s Law tan 𝑖𝐡 = πœ‡2 πœ‡1 ● If the light ray travels from one medium to another with refractive πœ‡1 and πœ‡2 respectively, then Brewster’s law becomes, Medium(πœ‡) 𝑖𝐡 π‘Ÿ
Relation between Critical Angle and Brewster Angle ∴ tan 𝑖𝐡 = 1 sin πœƒπΆ ● Brewster’s Law tan 𝑖𝐡 = πœ‡ 𝑖𝐡 = Brewster angle/polarising angle sin πœƒπΆ = 1 πœ‡ ● Critical Angle: 𝑖𝐡 = tanβˆ’1 1 sin πœƒπΆ And πœƒπΆ = sinβˆ’1 1 tan 𝑖𝐡 Medium(πœ‡) 𝑖𝐡 π‘Ÿ
T The polarizing angle of diamond is 67Β°. The critical angle of diamond is nearest to: [Given tan 67Β° = 2.36 ] To find: Solution: Critical angle Given: 𝑖𝐡 = 67Β° Critical angle is given by: πœƒπΆ = sinβˆ’1 1 2.36 1 2.36 < 1 2 β‡’ sinβˆ’1 1 2.36 < sinβˆ’1 1 2 β‡’ πœƒπΆ < 30Β° ∴ Out of the four option only 22Β° is less than 30Β° πœƒπΆ = sinβˆ’1 1 tan 𝑖𝐡 πœƒπΆ = sinβˆ’1 1 tan 67Β°
Scattering by Polarization ● If an unpolarized light gets scattered from air molecule, light perpendicular to original ray is plane polarized. ● If we draw a plane perpendicular to the incident light, then from every viewpoint on the plane we can see the plane polarized light. Air molecule Incident Light Observer Observer
Given: To find: Solution: 𝑖𝑝 = 60Β° Velocity of light in glass (𝑣𝑔) As reflected light is plane polarized, 𝑖𝑝 = 60∘ According to Brewster’s law, πœ‡ = tan 𝑖𝑝 = tan 60∘ = 3 As, πœ‡ = π‘£π‘Ž 𝑣𝑔 β‡’ 𝑣𝑔 = π‘£π‘Ž πœ‡ T 𝑣𝑔 = 3 Γ— 108 3 = 3 Γ— 108 π‘š/𝑠 A ray of light, travelling in air, is incident on a glass slab with angle of incidence 60Β°. It is found that the reflected ray is plane polarized. The velocity of light in the glass is: Air Plane polarized Glass (πœ‡) 90Β° 60Β°

Recommended

Ch 7 physical optics final
Ch 7 physical optics finalCh 7 physical optics final
Ch 7 physical optics finalanimesh samundh
Β 
Reflection & Refraction.pptx
Reflection & Refraction.pptxReflection & Refraction.pptx
Reflection & Refraction.pptxPaulBoro1
Β 
WAVES-converted.pdfgggghjnhdhbxbdbhdbdbbdhdhb
WAVES-converted.pdfgggghjnhdhbxbdbhdbdbbdhdhbWAVES-converted.pdfgggghjnhdhbxbdbhdbdbbdhdhb
WAVES-converted.pdfgggghjnhdhbxbdbhdbdbbdhdhbchettanagarwal
Β 
properties of light
properties of lightproperties of light
properties of lightKharen Adelan
Β 
A project on wave nature of light
A project on wave nature of lightA project on wave nature of light
A project on wave nature of lightKANNAN
Β 
04 UNIT-4 (WAVES) .pptx
04 UNIT-4 (WAVES) .pptx04 UNIT-4 (WAVES) .pptx
04 UNIT-4 (WAVES) .pptxFatimaAfzal56
Β 
6.wave_optics_1.ppt
6.wave_optics_1.ppt6.wave_optics_1.ppt
6.wave_optics_1.pptGaneshsaini17
Β 
Chapter 4a interference
Chapter 4a interferenceChapter 4a interference
Chapter 4a interferenceGabriel O'Brien
Β 

Recommended

Ch 7 physical optics final
Ch 7 physical optics finalCh 7 physical optics final
Ch 7 physical optics finalanimesh samundh
Β 
Reflection & Refraction.pptx
Reflection & Refraction.pptxReflection & Refraction.pptx
Reflection & Refraction.pptxPaulBoro1
Β 
WAVES-converted.pdfgggghjnhdhbxbdbhdbdbbdhdhb
WAVES-converted.pdfgggghjnhdhbxbdbhdbdbbdhdhbWAVES-converted.pdfgggghjnhdhbxbdbhdbdbbdhdhb
WAVES-converted.pdfgggghjnhdhbxbdbhdbdbbdhdhbchettanagarwal
Β 
properties of light
properties of lightproperties of light
properties of lightKharen Adelan
Β 
A project on wave nature of light
A project on wave nature of lightA project on wave nature of light
A project on wave nature of lightKANNAN
Β 
04 UNIT-4 (WAVES) .pptx
04 UNIT-4 (WAVES) .pptx04 UNIT-4 (WAVES) .pptx
04 UNIT-4 (WAVES) .pptxFatimaAfzal56
Β 
6.wave_optics_1.ppt
6.wave_optics_1.ppt6.wave_optics_1.ppt
6.wave_optics_1.pptGaneshsaini17
Β 
Chapter 4a interference
Chapter 4a interferenceChapter 4a interference
Chapter 4a interferenceGabriel O'Brien
Β 
Topic 4 Wave optics (English).pptx
Topic 4 Wave optics (English).pptxTopic 4 Wave optics (English).pptx
Topic 4 Wave optics (English).pptxSN20619PhangJianAn
Β 
Chapter 6 - Superposition of waves.pptx
Chapter 6 - Superposition of waves.pptxChapter 6 - Superposition of waves.pptx
Chapter 6 - Superposition of waves.pptxPooja M
Β 
Wave properties
Wave propertiesWave properties
Wave propertieswww.fixURscore.com
Β 
Wave motion
Wave motionWave motion
Wave motionchris lembalemba
Β 
Reflection__Refraction__and_Diffraction.ppt
Reflection__Refraction__and_Diffraction.pptReflection__Refraction__and_Diffraction.ppt
Reflection__Refraction__and_Diffraction.pptMuhammadShamraizDar
Β 
Reflection_Refraction_andDiffraction.ppt
Reflection_Refraction_andDiffraction.pptReflection_Refraction_andDiffraction.ppt
Reflection_Refraction_andDiffraction.pptJonalyn34
Β 
Interference of light waves
Interference of light wavesInterference of light waves
Interference of light wavesTahiraKhatoon3
Β 
class22A.ppt
class22A.pptclass22A.ppt
class22A.pptZiyaMahmad1
Β 
iii.pptx
iii.pptxiii.pptx
iii.pptxParasSundriyal2
Β 
SUBJECT: PHYSICS - Chapter 6 : Superposition of waves (CLASS XII - MAHARASH...
SUBJECT: PHYSICS - Chapter 6 : Superposition of waves (CLASS XII - MAHARASH... SUBJECT: PHYSICS - Chapter 6 : Superposition of waves (CLASS XII - MAHARASH...
SUBJECT: PHYSICS - Chapter 6 : Superposition of waves (CLASS XII - MAHARASH...Pooja M
Β 
2 interference and diffraction
2 interference and diffraction2 interference and diffraction
2 interference and diffractionEhab Hegazy
Β 
Reflection refraction
Reflection refractionReflection refraction
Reflection refractionSolo Hermelin
Β 
Optical Instrumentation 3. Interference
Optical Instrumentation 3. InterferenceOptical Instrumentation 3. Interference
Optical Instrumentation 3. InterferenceBurdwan University
Β 
wave pro.....
wave pro.....wave pro.....
wave pro.....Ajay Katiyar
Β 
Ondas Gravitacionales (en ingles)
Ondas Gravitacionales (en ingles)Ondas Gravitacionales (en ingles)
Ondas Gravitacionales (en ingles)Bishop's University, Canada
Β 
Ph 101-2
Ph 101-2Ph 101-2
Ph 101-2Chandan Singh
Β 
Engineering Physics Ch:20 WaveOptics.ppt
Engineering Physics Ch:20 WaveOptics.pptEngineering Physics Ch:20 WaveOptics.ppt
Engineering Physics Ch:20 WaveOptics.pptEngrNoumanMemon
Β 
Class 12th Physics wave optics ppt
Class 12th Physics wave optics pptClass 12th Physics wave optics ppt
Class 12th Physics wave optics pptArpit Meena
Β 
Diffraction of Light waves
Diffraction of Light wavesDiffraction of Light waves
Diffraction of Light wavesTahiraKhatoon3
Β 
Ray Optics Reduced New Syllabus.pdf
Ray Optics Reduced New Syllabus.pdfRay Optics Reduced New Syllabus.pdf
Ray Optics Reduced New Syllabus.pdfNileshkuBehera
Β 
Hire πŸ’• 9907093804 Hooghly Call Girls Service Call Girls Agency
Hire πŸ’• 9907093804 Hooghly Call Girls Service Call Girls AgencyHire πŸ’• 9907093804 Hooghly Call Girls Service Call Girls Agency
Hire πŸ’• 9907093804 Hooghly Call Girls Service Call Girls AgencySheetal Arora
Β 
Raman spectroscopy.pptx M Pharm, M Sc, Advanced Spectral Analysis
Raman spectroscopy.pptx M Pharm, M Sc, Advanced Spectral AnalysisRaman spectroscopy.pptx M Pharm, M Sc, Advanced Spectral Analysis
Raman spectroscopy.pptx M Pharm, M Sc, Advanced Spectral AnalysisDiwakar Mishra
Β 

More Related Content

Similar to Wave Optics Interference

Topic 4 Wave optics (English).pptx
Topic 4 Wave optics (English).pptxTopic 4 Wave optics (English).pptx
Topic 4 Wave optics (English).pptxSN20619PhangJianAn
Β 
Chapter 6 - Superposition of waves.pptx
Chapter 6 - Superposition of waves.pptxChapter 6 - Superposition of waves.pptx
Chapter 6 - Superposition of waves.pptxPooja M
Β 
Reflection__Refraction__and_Diffraction.ppt
Reflection__Refraction__and_Diffraction.pptReflection__Refraction__and_Diffraction.ppt
Reflection__Refraction__and_Diffraction.pptMuhammadShamraizDar
Β 
Reflection_Refraction_andDiffraction.ppt
Reflection_Refraction_andDiffraction.pptReflection_Refraction_andDiffraction.ppt
Reflection_Refraction_andDiffraction.pptJonalyn34
Β 
Interference of light waves
Interference of light wavesInterference of light waves
Interference of light wavesTahiraKhatoon3
Β 
class22A.ppt
class22A.pptclass22A.ppt
class22A.pptZiyaMahmad1
Β 
SUBJECT: PHYSICS - Chapter 6 : Superposition of waves (CLASS XII - MAHARASH...
SUBJECT: PHYSICS - Chapter 6 : Superposition of waves (CLASS XII - MAHARASH... SUBJECT: PHYSICS - Chapter 6 : Superposition of waves (CLASS XII - MAHARASH...
SUBJECT: PHYSICS - Chapter 6 : Superposition of waves (CLASS XII - MAHARASH...Pooja M
Β 
2 interference and diffraction
2 interference and diffraction2 interference and diffraction
2 interference and diffractionEhab Hegazy
Β 
Reflection refraction
Reflection refractionReflection refraction
Reflection refractionSolo Hermelin
Β 
Optical Instrumentation 3. Interference
Optical Instrumentation 3. InterferenceOptical Instrumentation 3. Interference
Optical Instrumentation 3. InterferenceBurdwan University
Β 
wave pro.....
wave pro.....wave pro.....
wave pro.....Ajay Katiyar
Β 
Engineering Physics Ch:20 WaveOptics.ppt
Engineering Physics Ch:20 WaveOptics.pptEngineering Physics Ch:20 WaveOptics.ppt
Engineering Physics Ch:20 WaveOptics.pptEngrNoumanMemon
Β 
Class 12th Physics wave optics ppt
Class 12th Physics wave optics pptClass 12th Physics wave optics ppt
Class 12th Physics wave optics pptArpit Meena
Β 
Diffraction of Light waves
Diffraction of Light wavesDiffraction of Light waves
Diffraction of Light wavesTahiraKhatoon3
Β 
Ray Optics Reduced New Syllabus.pdf
Ray Optics Reduced New Syllabus.pdfRay Optics Reduced New Syllabus.pdf
Ray Optics Reduced New Syllabus.pdfNileshkuBehera
Β 

Similar to Wave Optics Interference (20)

Topic 4 Wave optics (English).pptx
Topic 4 Wave optics (English).pptxTopic 4 Wave optics (English).pptx
Topic 4 Wave optics (English).pptx
Β 
Chapter 6 - Superposition of waves.pptx
Chapter 6 - Superposition of waves.pptxChapter 6 - Superposition of waves.pptx
Chapter 6 - Superposition of waves.pptx
Β 
Wave properties
Wave propertiesWave properties
Wave properties
Β 
Wave motion
Wave motionWave motion
Wave motion
Β 
Reflection__Refraction__and_Diffraction.ppt
Reflection__Refraction__and_Diffraction.pptReflection__Refraction__and_Diffraction.ppt
Reflection__Refraction__and_Diffraction.ppt
Β 
Reflection_Refraction_andDiffraction.ppt
Reflection_Refraction_andDiffraction.pptReflection_Refraction_andDiffraction.ppt
Reflection_Refraction_andDiffraction.ppt
Β 
Interference of light waves
Interference of light wavesInterference of light waves
Interference of light waves
Β 
class22A.ppt
class22A.pptclass22A.ppt
class22A.ppt
Β 
iii.pptx
iii.pptxiii.pptx
iii.pptx
Β 
SUBJECT: PHYSICS - Chapter 6 : Superposition of waves (CLASS XII - MAHARASH...
SUBJECT: PHYSICS - Chapter 6 : Superposition of waves (CLASS XII - MAHARASH... SUBJECT: PHYSICS - Chapter 6 : Superposition of waves (CLASS XII - MAHARASH...
SUBJECT: PHYSICS - Chapter 6 : Superposition of waves (CLASS XII - MAHARASH...
Β 
2 interference and diffraction
2 interference and diffraction2 interference and diffraction
2 interference and diffraction
Β 
Reflection refraction
Reflection refractionReflection refraction
Reflection refraction
Β 
Optical Instrumentation 3. Interference
Optical Instrumentation 3. InterferenceOptical Instrumentation 3. Interference
Optical Instrumentation 3. Interference
Β 
wave pro.....
wave pro.....wave pro.....
wave pro.....
Β 
Ondas Gravitacionales (en ingles)
Ondas Gravitacionales (en ingles)Ondas Gravitacionales (en ingles)
Ondas Gravitacionales (en ingles)
Β 
Ph 101-2
Ph 101-2Ph 101-2
Ph 101-2
Β 
Engineering Physics Ch:20 WaveOptics.ppt
Engineering Physics Ch:20 WaveOptics.pptEngineering Physics Ch:20 WaveOptics.ppt
Engineering Physics Ch:20 WaveOptics.ppt
Β 
Class 12th Physics wave optics ppt
Class 12th Physics wave optics pptClass 12th Physics wave optics ppt
Class 12th Physics wave optics ppt
Β 
Diffraction of Light waves
Diffraction of Light wavesDiffraction of Light waves
Diffraction of Light waves
Β 
Ray Optics Reduced New Syllabus.pdf
Ray Optics Reduced New Syllabus.pdfRay Optics Reduced New Syllabus.pdf
Ray Optics Reduced New Syllabus.pdf
Β 

Recently uploaded

Hire πŸ’• 9907093804 Hooghly Call Girls Service Call Girls Agency
Hire πŸ’• 9907093804 Hooghly Call Girls Service Call Girls AgencyHire πŸ’• 9907093804 Hooghly Call Girls Service Call Girls Agency
Hire πŸ’• 9907093804 Hooghly Call Girls Service Call Girls AgencySheetal Arora
Β 
Raman spectroscopy.pptx M Pharm, M Sc, Advanced Spectral Analysis
Raman spectroscopy.pptx M Pharm, M Sc, Advanced Spectral AnalysisRaman spectroscopy.pptx M Pharm, M Sc, Advanced Spectral Analysis
Raman spectroscopy.pptx M Pharm, M Sc, Advanced Spectral AnalysisDiwakar Mishra
Β 
Lucknow πŸ’‹ Russian Call Girls Lucknow Finest Escorts Service 8923113531 Availa...
Lucknow πŸ’‹ Russian Call Girls Lucknow Finest Escorts Service 8923113531 Availa...Lucknow πŸ’‹ Russian Call Girls Lucknow Finest Escorts Service 8923113531 Availa...
Lucknow πŸ’‹ Russian Call Girls Lucknow Finest Escorts Service 8923113531 Availa...anilsa9823
Β 
Pests of cotton_Sucking_Pests_Dr.UPR.pdf
Pests of cotton_Sucking_Pests_Dr.UPR.pdfPests of cotton_Sucking_Pests_Dr.UPR.pdf
Pests of cotton_Sucking_Pests_Dr.UPR.pdfPirithiRaju
Β 
GFP in rDNA Technology (Biotechnology).pptx
GFP in rDNA Technology (Biotechnology).pptxGFP in rDNA Technology (Biotechnology).pptx
GFP in rDNA Technology (Biotechnology).pptxAleenaTreesaSaji
Β 
Pests of cotton_Borer_Pests_Binomics_Dr.UPR.pdf
Pests of cotton_Borer_Pests_Binomics_Dr.UPR.pdfPests of cotton_Borer_Pests_Binomics_Dr.UPR.pdf
Pests of cotton_Borer_Pests_Binomics_Dr.UPR.pdfPirithiRaju
Β 
Biopesticide (2).pptx .This slides helps to know the different types of biop...
Biopesticide (2).pptx .This slides helps to know the different types of biop...Biopesticide (2).pptx .This slides helps to know the different types of biop...
Biopesticide (2).pptx .This slides helps to know the different types of biop...RohitNehra6
Β 
Recombinant DNA technology (Immunological screening)
Recombinant DNA technology (Immunological screening)Recombinant DNA technology (Immunological screening)
Recombinant DNA technology (Immunological screening)PraveenaKalaiselvan1
Β 
Green chemistry and Sustainable development.pptx
Green chemistry and Sustainable development.pptxGreen chemistry and Sustainable development.pptx
Green chemistry and Sustainable development.pptxRajatChauhan518211
Β 
Hubble Asteroid Hunter III. Physical properties of newly found asteroids
Hubble Asteroid Hunter III. Physical properties of newly found asteroidsHubble Asteroid Hunter III. Physical properties of newly found asteroids
Hubble Asteroid Hunter III. Physical properties of newly found asteroidsSΓ©rgio Sacani
Β 
Recombination DNA Technology (Nucleic Acid Hybridization )
Recombination DNA Technology (Nucleic Acid Hybridization )Recombination DNA Technology (Nucleic Acid Hybridization )
Recombination DNA Technology (Nucleic Acid Hybridization )aarthirajkumar25
Β 
Stunning βž₯8448380779β–» Call Girls In Panchshil Enclave Delhi NCR
Stunning βž₯8448380779β–» Call Girls In Panchshil Enclave Delhi NCRStunning βž₯8448380779β–» Call Girls In Panchshil Enclave Delhi NCR
Stunning βž₯8448380779β–» Call Girls In Panchshil Enclave Delhi NCRDelhi Call girls
Β 
GBSN - Biochemistry (Unit 1)
GBSN - Biochemistry (Unit 1)GBSN - Biochemistry (Unit 1)
GBSN - Biochemistry (Unit 1)Areesha Ahmad
Β 
Botany 4th semester file By Sumit Kumar yadav.pdf
Botany 4th semester file By Sumit Kumar yadav.pdfBotany 4th semester file By Sumit Kumar yadav.pdf
Botany 4th semester file By Sumit Kumar yadav.pdfSumit Kumar yadav
Β 
Nightside clouds and disequilibrium chemistry on the hot Jupiter WASP-43b
Nightside clouds and disequilibrium chemistry on the hot Jupiter WASP-43bNightside clouds and disequilibrium chemistry on the hot Jupiter WASP-43b
Nightside clouds and disequilibrium chemistry on the hot Jupiter WASP-43bSΓ©rgio Sacani
Β 
Natural Polymer Based Nanomaterials
Natural Polymer Based NanomaterialsNatural Polymer Based Nanomaterials
Natural Polymer Based NanomaterialsAArockiyaNisha
Β 
PossibleEoarcheanRecordsoftheGeomagneticFieldPreservedintheIsuaSupracrustalBe...
PossibleEoarcheanRecordsoftheGeomagneticFieldPreservedintheIsuaSupracrustalBe...PossibleEoarcheanRecordsoftheGeomagneticFieldPreservedintheIsuaSupracrustalBe...
PossibleEoarcheanRecordsoftheGeomagneticFieldPreservedintheIsuaSupracrustalBe...SΓ©rgio Sacani
Β 
Nanoparticles synthesis and characterization​ ​
Nanoparticles synthesis and characterization​ ​Nanoparticles synthesis and characterization​ ​
Nanoparticles synthesis and characterization​ ​kaibalyasahoo82800
Β 

Recently uploaded (20)

Hire πŸ’• 9907093804 Hooghly Call Girls Service Call Girls Agency
Hire πŸ’• 9907093804 Hooghly Call Girls Service Call Girls AgencyHire πŸ’• 9907093804 Hooghly Call Girls Service Call Girls Agency
Hire πŸ’• 9907093804 Hooghly Call Girls Service Call Girls Agency
Β 
Raman spectroscopy.pptx M Pharm, M Sc, Advanced Spectral Analysis
Raman spectroscopy.pptx M Pharm, M Sc, Advanced Spectral AnalysisRaman spectroscopy.pptx M Pharm, M Sc, Advanced Spectral Analysis
Raman spectroscopy.pptx M Pharm, M Sc, Advanced Spectral Analysis
Β 
Lucknow πŸ’‹ Russian Call Girls Lucknow Finest Escorts Service 8923113531 Availa...
Lucknow πŸ’‹ Russian Call Girls Lucknow Finest Escorts Service 8923113531 Availa...Lucknow πŸ’‹ Russian Call Girls Lucknow Finest Escorts Service 8923113531 Availa...
Lucknow πŸ’‹ Russian Call Girls Lucknow Finest Escorts Service 8923113531 Availa...
Β 
Pests of cotton_Sucking_Pests_Dr.UPR.pdf
Pests of cotton_Sucking_Pests_Dr.UPR.pdfPests of cotton_Sucking_Pests_Dr.UPR.pdf
Pests of cotton_Sucking_Pests_Dr.UPR.pdf
Β 
GFP in rDNA Technology (Biotechnology).pptx
GFP in rDNA Technology (Biotechnology).pptxGFP in rDNA Technology (Biotechnology).pptx
GFP in rDNA Technology (Biotechnology).pptx
Β 
Pests of cotton_Borer_Pests_Binomics_Dr.UPR.pdf
Pests of cotton_Borer_Pests_Binomics_Dr.UPR.pdfPests of cotton_Borer_Pests_Binomics_Dr.UPR.pdf
Pests of cotton_Borer_Pests_Binomics_Dr.UPR.pdf
Β 
Biopesticide (2).pptx .This slides helps to know the different types of biop...
Biopesticide (2).pptx .This slides helps to know the different types of biop...Biopesticide (2).pptx .This slides helps to know the different types of biop...
Biopesticide (2).pptx .This slides helps to know the different types of biop...
Β 
Recombinant DNA technology (Immunological screening)
Recombinant DNA technology (Immunological screening)Recombinant DNA technology (Immunological screening)
Recombinant DNA technology (Immunological screening)
Β 
Green chemistry and Sustainable development.pptx
Green chemistry and Sustainable development.pptxGreen chemistry and Sustainable development.pptx
Green chemistry and Sustainable development.pptx
Β 
Hubble Asteroid Hunter III. Physical properties of newly found asteroids
Hubble Asteroid Hunter III. Physical properties of newly found asteroidsHubble Asteroid Hunter III. Physical properties of newly found asteroids
Hubble Asteroid Hunter III. Physical properties of newly found asteroids
Β 
The Philosophy of Science
The Philosophy of ScienceThe Philosophy of Science
The Philosophy of Science
Β 
Recombination DNA Technology (Nucleic Acid Hybridization )
Recombination DNA Technology (Nucleic Acid Hybridization )Recombination DNA Technology (Nucleic Acid Hybridization )
Recombination DNA Technology (Nucleic Acid Hybridization )
Β 
Stunning βž₯8448380779β–» Call Girls In Panchshil Enclave Delhi NCR
Stunning βž₯8448380779β–» Call Girls In Panchshil Enclave Delhi NCRStunning βž₯8448380779β–» Call Girls In Panchshil Enclave Delhi NCR
Stunning βž₯8448380779β–» Call Girls In Panchshil Enclave Delhi NCR
Β 
GBSN - Biochemistry (Unit 1)
GBSN - Biochemistry (Unit 1)GBSN - Biochemistry (Unit 1)
GBSN - Biochemistry (Unit 1)
Β 
Botany 4th semester file By Sumit Kumar yadav.pdf
Botany 4th semester file By Sumit Kumar yadav.pdfBotany 4th semester file By Sumit Kumar yadav.pdf
Botany 4th semester file By Sumit Kumar yadav.pdf
Β 
Nightside clouds and disequilibrium chemistry on the hot Jupiter WASP-43b
Nightside clouds and disequilibrium chemistry on the hot Jupiter WASP-43bNightside clouds and disequilibrium chemistry on the hot Jupiter WASP-43b
Nightside clouds and disequilibrium chemistry on the hot Jupiter WASP-43b
Β 
9953056974 Young Call Girls In Mahavir enclave Indian Quality Escort service
9953056974 Young Call Girls In Mahavir enclave Indian Quality Escort service9953056974 Young Call Girls In Mahavir enclave Indian Quality Escort service
9953056974 Young Call Girls In Mahavir enclave Indian Quality Escort service
Β 
Natural Polymer Based Nanomaterials
Natural Polymer Based NanomaterialsNatural Polymer Based Nanomaterials
Natural Polymer Based Nanomaterials
Β 
PossibleEoarcheanRecordsoftheGeomagneticFieldPreservedintheIsuaSupracrustalBe...
PossibleEoarcheanRecordsoftheGeomagneticFieldPreservedintheIsuaSupracrustalBe...PossibleEoarcheanRecordsoftheGeomagneticFieldPreservedintheIsuaSupracrustalBe...
PossibleEoarcheanRecordsoftheGeomagneticFieldPreservedintheIsuaSupracrustalBe...
Β 
Nanoparticles synthesis and characterization​ ​
Nanoparticles synthesis and characterization​ ​Nanoparticles synthesis and characterization​ ​
Nanoparticles synthesis and characterization​ ​
Β 

Wave Optics Interference

  • 2. Does Light always exhibit Rectilinear Propagation?
  • 3. Explanation for bending of light Wave theory of light Christiaan Huygens Wavefront Huygens’ Principle Huygens Wave theory of light
  • 4. ο‚— All points on a particular circle will oscillate with the same phase. ο‚— Wavefront is the locus of all points at which the wave disturbance is in the same phase. Wavefront ο‚— In 3𝐷, for a point source, the wavefronts are spherically symmetric. ο‚— Point source creates wave traveling in all directions according to equation. 𝑦 = 𝐴sin(πœ”π‘‘ Β± π‘˜π‘₯) ο‚— Points 𝐴, 𝐡, 𝐢 have same phase πœ”π‘‘ Β± π‘˜π‘₯1.
  • 5. The direction of propagation of light wave is perpendicular to the wavefront at any given point. Direction of Propagation of Light Spherical Wavefront
  • 6. ο‚— Time taken by each ray to propagate from one wavefront to the next wavefront in any medium remains same. ο‚— Hence, distance covered by the light in each time interval also changes. ⟹ Speed of light changes. ο‚— Refractive index, 𝑛 = 𝑐 𝑣 When medium changes, 𝑛1 𝑛2 = 𝑣2 𝑣1 Properties of Wavefront πœ† ο‚— Distance between two consecutive wavefronts = πœ†.
  • 7. ο‚— Spherical wavefront is observed for a point source. β€’ Direction of propagation of light: Radially outward and perpendicular to the wavefront at any given point. Spherical Wavefront
  • 8. β€’ Planar wavefronts are observed when the source is at infinity. β€’ Direction of propagation of light: Perpendicular to the plane. Planar Wavefront
  • 9. β€’ Cylindrical wavefront is observed for a line source. β€’ Direction of propagation of light: Radially outward and perpendicular to the wavefront at any given point. Cylindrical Wavefront
  • 10. S If the distance between the wavefronts in the medium with refractive index 𝑛2 is 𝑑2, then what will be the distance 𝑑1 between the wavefronts in the medium with refractive index 𝑛1? 𝑛 = 𝑐 𝑣 We known that, ΰ΅— 𝑛2 𝑛1 = ΰ΅— 𝑣1 𝑣2 πœ† = ΰ΅— 𝑣 𝑓 β‡’ β‡’ ΰ΅— πœ†1 πœ†2 = ΰ΅— 𝑣1 𝑣2 = ΰ΅— 𝑛2 𝑛1 Here, πœ†1 = 𝑑1, and πœ†2 = 𝑑2 Hence, 𝑑1 𝑑2 = 𝑛2 𝑛1 β‡’ 𝑑1 = 𝑑2 𝑛2 𝑛1
  • 11. ο‚— Every point on the wavefront acts as a point source called secondary wave source and generates secondary wavelets. ο‚— The common tangent to the secondary wavelets in the forward direction gives the secondary wavefront. ο‚— Intensity is maximum in forward direction and zero in backward direction. Secondary Wavelets Secondary Wavefront Primary Wavefront Primary Source Huygens’ Principle
  • 12. Shape of Wavefront Object at infinity for a convex lens Object at infinity for a concave lens
  • 13. Object at infinity for a concave mirror Shape of Wavefront Object at infinity for a prism
  • 14. Reflection of a plane wave by a plane surface Huygens’ Principle - Law of Reflection Reflecting plane surface 𝐴 𝐡 𝐡𝐢 = 𝑣𝑑 𝐢 𝐷 𝐷𝐴 = 𝑣𝑑 𝑑 = 0 𝑖 𝑑 Incident Wavefront 𝐴𝐡 at 𝑑 = 0 Reflected Wavefront 𝐢𝐷 at time = 𝑑 Secondary Wavelet From 𝐴
  • 15. Reflection of a plane wave by a plane surface Huygens’ Principle - Law of Reflection ∠𝐷𝐢𝐴 = π‘Ÿ β†’ Angle of reflection ∠𝐡𝐴𝐢 = 𝑖 β†’ Angle of incidence ο‚— Angle of incidence is the angle between the incident wavefront and the reflecting surface. ο‚— Angle of reflection is the angle between the reflected wavefront and the reflecting surface. 𝐴 𝐢 𝐷 π‘Ÿ 𝐡 𝐢 𝐴 𝑖 ∴ βˆ†π΄π·πΆ β‰… βˆ†πΆπ΅π΄ ⟹ βˆ π‘– = βˆ π‘Ÿ From βˆ†π΄π·πΆ and βˆ†πΆπ΅π΄: ∠𝐴𝐷𝐢 = ∠𝐴𝐡𝐢 = 90Β° 𝐴𝐷 = 𝐡𝐢 = 𝑣𝑑 𝐴𝐢 = 𝐴𝐢 [𝑅. 𝐻. 𝑆 Congruency]
  • 16. From βˆ†π΄π·πΆ: From βˆ†π΄π΅πΆ: sin 𝑖 = 𝐡𝐢 𝐴𝐢 = 𝑣1 𝑑 𝐴𝐢 sin π‘Ÿ = 𝐴𝐷 𝐴𝐢 = 𝑣2 𝑑 𝐴𝐢 sin 𝑖 sin π‘Ÿ = 𝑣1 𝑣2 = 𝑛2 𝑛1 = Constant Huygens’ Principle - Law of Refraction Medium 𝑛2 𝐴 𝐡 𝑣1𝑑 𝐢 𝐷 𝑖 π‘Ÿ 𝑑 = 0 π‘Ÿ 𝑖 Incident Wavefront 𝐴𝐡 at 𝑑 = 0 Refracted Wavefront 𝐢𝐷 at 𝑑 = 𝑑 Secondary Wavelet From A 𝑣2𝑑 π‘Ÿ Medium 𝑛1 𝑛1 < 𝑛2
  • 17. B A D C Solution: The direction of propagation β†’ ΖΈ 𝑖 ⟹ The direction of wavefront β†’ βŠ₯ π‘‘π‘œ ΖΈ 𝑖 In the given options, the plane βŠ₯ to ΖΈ 𝑖 is represented by π‘₯ = 𝑐. π‘₯ = 𝑐 𝑦 = 𝑐 𝑧 = 𝑐 π‘₯ + 𝑦 + 𝑧 = 𝑐 Light waves travel in vacuum, along the π‘₯ βˆ’ axis. Which of the following may represent the wavefronts?
  • 18. Hole Screen Hole Screen Light behaves like a ray (i.e., exhibits rectilinear propagation) for: Obstacle dimensions β‰«πœ†π‘™π‘–π‘”β„Žπ‘‘ Obstacle dimensions β‰ˆ πœ†π‘™π‘–π‘”β„Žπ‘‘ Light behaves like a wave for: Does Light always exhibit Rectilinear Propagation?
  • 19. OPTICS Study of the behavior and properties of light GEOMETRICAL OPTICS Explains refraction and reflection, but not interference, diffraction and polarization WAVE OPTICS Explains reflection and refraction, as well as interference, diffraction and polarization
  • 20. Interference is the phenomenon in which two waves superpose to form the resultant wave of lower, higher or same amplitude. Interference
  • 21. β€œWhen two or more waves cross at a point, the displacement at that point is equal to the vector sum of the displacements of individual waves” Τ¦ 𝑦𝑛𝑒𝑑 = Τ¦ 𝑦1 + Τ¦ 𝑦2 + Τ¦ 𝑦3 … … + Τ¦ 𝑦𝑛 𝑦 π‘₯ 𝑂 𝑦1 𝑦2 𝑦𝑛𝑒𝑑 Superposition Principle
  • 22. Case 1: When two crest meet Case 2: When crest and trough meet Resultant of Waves
  • 23. Phase Difference and Path Difference ● Path difference (βˆ†π‘₯): Difference in the path traversed by the two waves. ● Phase difference (𝛿): Difference in the phase angle of the two waves. We know that, For a path difference πœ†, phase difference = 2πœ‹ So, for path difference βˆ†π‘₯, phase difference = 2πœ‹ πœ† βˆ†π‘₯. 𝛿 = 2πœ‹ πœ† βˆ†π‘₯ Source 𝑆1 Source 𝑆2 𝑦1 = 𝐴1 sin πœ”π‘‘ + π‘˜π‘₯ 𝑦2 = 𝐴2 sin πœ”π‘‘ + π‘˜π‘₯ + 𝛿 𝑦𝑛𝑒𝑑 = 𝑦1 + 𝑦2 𝑃
  • 24. Combination of Waves 𝑦𝑛𝑒𝑑 = 𝐴1 sin πœ”π‘‘ + π‘˜π‘₯ + 𝐴2 sin πœ”π‘‘ + π‘˜π‘₯ + 𝛿 𝐴𝑛𝑒𝑑 = 𝐴1 2 + 𝐴2 2 + 2𝐴1𝐴2 cos 𝛿 tan 𝛼 = 𝐴2 sin 𝛿 𝐴1 + 𝐴2 cos 𝛿 𝑦𝑛𝑒𝑑 = 𝐴𝑛𝑒𝑑 sin πœ”π‘‘ + π‘˜π‘₯ + 𝛼 If 𝐴1 = 𝐴2 = 𝐴 𝐴𝑛𝑒𝑑 = 2𝐴 cos 𝛿 2
  • 25. Constructive Interference Interference that produces maximum possible amplitude (or maximum intensity) is called constructive interference. 𝑃 2πœ† 3πœ† ● Path Difference = 3πœ† βˆ’ 2πœ† = πœ† 𝐴𝑛𝑒𝑑 = 𝐴1 2 + 𝐴2 2 + 2𝐴1𝐴2 cos 𝛿 For maximum amplitude: cos 𝛿 = 1 β‡’ 𝛿 = 2π‘›πœ‹ 𝐴 = π΄π‘šπ‘Žπ‘₯ = 𝐴1 + 𝐴2 2π‘›πœ‹ = 2πœ‹ πœ† Γ— (βˆ†π‘₯) Path difference = βˆ†π‘₯ = π‘›πœ† 𝛿 = 2πœ‹ πœ† βˆ†π‘₯ Constructive Interference
  • 26. ≑ π‘₯ 𝑦 𝑦𝑛𝑒𝑑 = 2𝐴 sin(πœ”π‘‘ + π‘˜π‘₯) + Constructive interference occurs when the crest and trough of one wave overlaps with the crest and trough of another wave. Constructive Interference π‘₯ 𝑦 𝑦1 = 𝐴 sin(πœ”π‘‘ + π‘˜π‘₯) 𝑂 π‘₯ 𝑦 𝑦2 = 𝐴 sin(πœ”π‘‘ + π‘˜π‘₯) 𝑂
  • 27. Destructive Interference Interference that produces minimum possible amplitude (or minimum intensity) is called destructive interference. 𝐴𝑛𝑒𝑑 = 𝐴1 2 + 𝐴2 2 + 2𝐴1𝐴2 cos 𝛿 𝑃 7.25πœ† 9.75πœ† For minimum amplitude: cos 𝛿 = βˆ’1 β‡’ 𝛿 = (2𝑛 + 1)πœ‹ 𝐴 = π΄π‘šπ‘–π‘› = 𝐴1 βˆ’ 𝐴2 (2𝑛 + 1)πœ‹ = 2πœ‹ πœ† (βˆ†π‘₯) Path difference = 𝛿 = 2πœ‹ πœ† βˆ†π‘₯ βˆ†π‘₯ = 2𝑛 + 1 πœ† 2 ● Path Difference = 9.75πœ† βˆ’ 7.25πœ† = 2.5πœ† Destructive Interference
  • 28. π‘₯ 𝑦 𝑦1 = 𝐴 sin(πœ”π‘‘ + π‘˜π‘₯) Destructive interference is resulted when the crest of one wave overlaps with the trough of another wave. ≑ π‘₯ 𝑦 𝑂 𝑦 = 0 + π‘₯ 𝑦 𝑦2 = 𝐴 sin(πœ”π‘‘ + π‘˜π‘₯ + πœ‹) Destructive Interference 𝑂 𝑂
  • 29. T Two-point light sources 𝑆1 and 𝑆2 are separated by a distance of 4.2πœ†. If an observer standing at the centre 𝐢 of the two sources starts moving towards 𝑆2, then find the minimum distance travelled by the observer to meet the first maxima. 𝑆1 𝑆2 2.1πœ† 𝐢 𝑃 βˆ†π‘₯ = 0 π‘₯ 2.1πœ† βˆ†π‘₯ = πœ† Solution: For maxima at 𝑃: βˆ†π‘₯ = πœ† β‡’ 𝑆1𝑃 βˆ’ 𝑆2𝑃 = πœ† β‡’ (2.1πœ† + π‘₯) βˆ’ (2.1πœ† βˆ’ π‘₯) = πœ† β‡’ 2π‘₯ = πœ† π‘₯ = 0.5πœ†
  • 30. Coherent and Incoherent Sources Coherent Waves Incoherent Waves Coherent Sources ● Same wavelength ● Same frequency ● Constant phase difference Incoherent Sources ● Different wavelength ● Different frequency ● Varying phase difference Note: We will always consider coherent sources in our discussion.
  • 31. Young’s Double Slit Experiment Particle nature of light Wave nature of light
  • 32. According to Huygens’ principle, the sources 𝑆1 and 𝑆2 will behave as independent sources. Young’s Double Slit Experiment 𝐷 ● Light source must be monochromatic. ● Sources 𝑆1 and 𝑆2 must be coherent. ● Width of the slit is comparable to the wavelength of light. ● Waves coming from sources 𝑆1 and 𝑆2 will interference and obtain different interference pattern on the screen.
  • 33. Young’s Double Slit Experiment β€’ 𝑆𝑆1 = 𝑆𝑆2 𝐷 (∴ No path difference till slits.) 𝑑 β€’ ′𝑑′ is distance between slits. β€’ 𝑂𝐡 is the central line. 𝑂 𝐡 β€’ ′𝐷′ is distance between screen and slits plane. πœƒ 𝑃 𝑦 β€’ Consider point 𝑃 at a distance 𝑦 from central line 𝑂𝐡. β€’ βˆ π‘ƒπ‘‚π΅ = πœƒ. β€’ 𝑆2𝑃 > 𝑆1𝑃 β†’ 𝑆2𝑃 βˆ’ 𝑆1𝑃 = βˆ†π‘₯ β€’ βˆ†π‘₯ is the path difference.
  • 34. Young’s Double Slit Experiment 𝐷 𝑑 𝑂 𝐡 πœƒ 𝑃 𝑦 𝑑/2 𝑑/2 ● Path difference at any general point 𝑃, βˆ†π‘₯ = 𝑆2𝑃 βˆ’ 𝑆1𝑃 βˆ†π‘₯ = 𝑦 + 𝑑 2 2 + 𝐷2 βˆ’ 𝑦 βˆ’ 𝑑 2 2 + 𝐷2 Approximation 1: 𝐷 ≫ 𝑑 Approximation 2: πœƒ is verysmall βˆ†π‘₯ β‰ˆ 𝑦𝑑 𝐷
  • 35. Young’s Double Slit Experiment βˆ†π‘₯ = 𝑆2𝐴 = 𝑑 sin πœƒ πœƒ is verysmall β†’ sin πœƒ β‰ˆ tan πœƒ βˆ†π‘₯ β‰ˆ 𝑑 tan πœƒ = 𝑑𝑦 𝐷 βˆ†π‘₯ β‰ˆ 𝑦𝑑 𝐷
  • 36. ● For Maxima (Constructive interference): βˆ†π‘₯ = π‘›πœ† 𝑦 = π‘›πœ†π· 𝑑 Where, 𝑛 = 0, Β±1, Β±2, Β±3, … β‡’ 𝑑𝑦 𝐷 = π‘›πœ† 𝑛 = 0 correspondstothecentralmaxima. 𝑛 = Β±1 correspondtothe1𝑠𝑑 maxima. 𝑛 = Β±2 correspondtothe2𝑛𝑑 maxima. 𝑃 𝐡 𝑑 𝑂 𝑦 𝑆1 𝑆2 𝐷 βˆ†π‘₯ Condition for Constructive Interference β†’ 𝑦 = 0 β†’ 𝑦 = Β± πœ†π· 𝑑 β†’ 𝑦 = Β± 2πœ†π· 𝑑
  • 37. ● For Minima (Destructive interference): βˆ†π‘₯ = 𝑛 + 1 2 πœ† β‡’ 𝑑𝑦 𝐷 = 𝑛 + 1 2 πœ† 𝑦 = 𝑛 + 1 2 πœ†π· 𝑑 𝑛 = 0,βˆ’1 correspondtothe1𝑠𝑑 minima. 𝑛 = 1,βˆ’2 correspondtothe2𝑛𝑑 minima. 𝑃 𝐡 𝑑 𝑂 𝑦 𝑆1 𝑆2 𝐷 βˆ†π‘₯ Condition for Destructive Interference Where, 𝑛 = 0, Β±1, Β±2, Β±3, … β†’ 𝑦 = Β± πœ†π· 2𝑑 β†’ 𝑦 = Β± 3πœ†π· 2𝑑
  • 38. In YDSE, white light is passed through the double slit and interference pattern is observed on a screen 2.5 π‘š away. The separation between the slits is 0.5 π‘šπ‘š. The first violet and red maxima are formed at distances of 2 π‘šπ‘š and 3.5 π‘šπ‘š away from the central white maxima, respectively. The wavelengths of red and violet light, respectively, are: Given: To find: Solution: 𝐷 = 2.5 π‘š, 𝑑 = 0.5 π‘šπ‘š, π‘¦π‘£π‘–π‘œπ‘™π‘’π‘‘ = 2 π‘šπ‘š, π‘¦π‘Ÿπ‘’π‘‘ = 3.5 π‘šπ‘š πœ†π‘£π‘–π‘œπ‘™π‘’π‘‘, πœ†π‘Ÿπ‘’π‘‘ Distance of first maxima from the central maxima is given by: For violet light, 𝑦 = 2 π‘šπ‘š β‡’ 2 Γ— 10βˆ’3 = 2.5 Γ— πœ†π‘£π‘–π‘œπ‘™π‘’π‘‘ 0.5 Γ— 10βˆ’3 β‡’ πœ†π‘£π‘–π‘œπ‘™π‘’π‘‘ = 2 5 Γ— 10βˆ’6 π‘š 𝑦 = π·πœ† 𝑑 For red light, 𝑦 = 3.5 π‘šπ‘š β‡’ 3.5 Γ— 10βˆ’3 = 2.5 Γ— πœ†π‘Ÿπ‘’π‘‘ 0.5 Γ— 10βˆ’3 β‡’ πœ†π‘Ÿπ‘’π‘‘= 0.7 Γ— 10βˆ’6 π‘š πœ†π‘Ÿπ‘’π‘‘ = 700 π‘›π‘š πœ†π‘£π‘–π‘œπ‘™π‘’π‘‘ = 400 π‘›π‘š
  • 39. Resultant Amplitude of the Wave 𝐸1 = 𝐸01 sin π‘˜π‘₯ βˆ’ πœ”π‘‘ 𝐸2 = 𝐸02 sin π‘˜π‘₯ βˆ’ πœ”π‘‘ + 𝛿 𝐸01, 𝐸02 π‘œπ‘Ÿ 𝐴01, 𝐴02 β†’ Electric Field Amplitude. Electric field of the wave from 𝑆1: Electric field of the wave from 𝑆2: 𝐸0 2 = 𝐸01 2 + 𝐸02 2 + 2𝐸01𝐸02 cos 𝛿 𝐸𝑛𝑒𝑑 = 𝐸1 + 𝐸2 = 𝐸0 sin π‘˜π‘₯ βˆ’ πœ”π‘‘ + 𝛿 Net electric field after the interference: tan πœ€ = 𝐸02 sin 𝛿 𝐸01 + 𝐸02 cos 𝛿
  • 40. Intensity of Waves 𝐼 = 2πœ‹2 𝑓2 πœŒπ‘£πΈ0 2 𝐼 ∝ 𝐸0 2 𝐼𝑛𝑒𝑑 = 𝐼1 + 𝐼2 + 2 𝐼1 𝐼2 cos 𝛿 𝐸0 2 = 𝐸01 2 + 𝐸02 2 + 2𝐸01𝐸02 cos 𝛿 𝑃 𝐡 𝑑 𝑂 𝑦 𝑆1 𝑆2 𝐷 𝐸2 𝐸1
  • 41. Intensity for Identical Waves 𝐼𝑛𝑒𝑑 = 4𝐼 cos2 𝛿 2 𝐼𝑛𝑒𝑑 = 𝐼1 + 𝐼2 + 2 𝐼1 𝐼2 cos 𝛿 ● For identical slits: 𝐼1 = 𝐼2 = 𝐼 𝐼𝑛𝑒𝑑 = 𝐼 + 𝐼 + 2 𝐼 𝐼 cos 𝛿 𝐼𝑛𝑒𝑑 = 2𝐼 + 2𝐼 cos 𝛿 = 2𝐼(1 + cos 𝛿) 𝐼𝑛𝑒𝑑 = 2𝐼 Γ— 2 cos2 𝛿 2 𝑃 𝐡 𝑑 𝑂 𝑦 𝑆1 𝑆2 𝐷 𝐼2 𝐼1
  • 42. ● For constructive interference: (π‘€π‘Žπ‘₯π‘–π‘šπ‘Ž) Intensity of Waves ● For destructive interference: (π‘€π‘–π‘›π‘–π‘šπ‘Ž) 𝐸0 π‘šπ‘Žπ‘₯ = 𝐸01 + 𝐸02 𝐸0 π‘šπ‘–π‘› = 𝐸01 βˆ’ 𝐸02 𝐼 ∝ 𝐸0 2 β‡’ πΌπ‘šπ‘Žπ‘₯ πΌπ‘šπ‘–π‘› = 𝐸0 π‘šπ‘Žπ‘₯ 𝐸0 π‘šπ‘–π‘› 2 πΌπ‘šπ‘Žπ‘₯ πΌπ‘šπ‘–π‘› = 𝐸01 + 𝐸02 𝐸01 βˆ’ 𝐸02 2 ● When cos 𝛿 = 1, 𝛿 = 2π‘›πœ‹ (π‘€π‘Žπ‘₯π‘–π‘šπ‘Ž) ● When cos 𝛿 = βˆ’1, 𝛿 = 2𝑛 + 1 πœ‹ (π‘€π‘–π‘›π‘–π‘šπ‘Ž) 𝐼𝑛𝑒𝑑 = 𝐼1 + 𝐼2 + 2 𝐼1 𝐼2 β‡’ 𝐼1 + 𝐼2 2 𝐼𝑛𝑒𝑑 = 𝐼1 + 𝐼2 βˆ’ 2 𝐼1 𝐼2 β‡’ 𝐼1 βˆ’ 𝐼2 2 𝐼 = 𝐼1 + 𝐼2 2 , constructive interference 𝐼1 βˆ’ 𝐼2 2 , destructive interference
  • 43. Two coherent sources produce waves of different intensities which interfere. After interference, the ratio of the maximum intensity to the minimum intensity is 16. The intensity of the waves are in the ratio: 𝐼1 𝐼2 = 25 9 Solution: Given: To find: πΌπ‘šπ‘Žπ‘₯ πΌπ‘šπ‘–π‘› = 16 𝐼1 𝐼2 πΌπ‘šπ‘Žπ‘₯ πΌπ‘šπ‘–π‘› = 16 β‡’ 𝐼1 + 𝐼2 𝐼1 βˆ’ 𝐼2 2 = 16 1 β‡’ 𝐼1 + 𝐼2 = 4 𝐼1 βˆ’ 4 𝐼2 β‡’ 3 𝐼1 = 5 𝐼2 β‡’ 𝐼1 𝐼2 = 5 3 Ratio of intensities: 𝐼1 𝐼2 = 5 3 2
  • 44. Intensity Variation 𝑦 = π‘›πœ†π· 𝑑 𝑦 = 𝑛 + 1 2 πœ†π· 𝑑 ● Constructive Interference: ● Destructive Interference: 𝑦 1𝑠𝑑 order maxima 2nd order maxima Central maxima 1𝑠𝑑 order minima 2𝑠𝑑 order minima 𝑦 = Β± πœ†π· 𝑑 , Β± 2πœ†π· 𝑑 , Β± 3πœ†π· 𝑑 . . . . . . . 𝑦 = Β± πœ†π· 2𝑑 , Β± 3πœ†π· 2𝑑 , Β± 5πœ†π· 2𝑑 . . . . . . . 𝐷 𝑂 𝑆1 𝑆2 𝑑 𝐼0 𝐼0
  • 45. Intensity Variation 𝐼 = 4𝐼0 cos2 𝛿 2 ● Intensity at any point: ● πΌπ‘šπ‘Žπ‘₯ = 4𝐼0, πΌπ‘šπ‘–π‘› = 0 𝛿 = Phase difference between the two waves from 𝑆1 and 𝑆2. 𝐼0 𝐼0 4𝐼0 𝐼 = 0 𝐼 = 0 𝐼 = 0 𝐼 = 0 𝐷 𝑂 𝑆1 𝑆2 𝑑 4𝐼0 4𝐼0 4𝐼0 4𝐼0
  • 46. Shape of Fringes on Screen If we replace slits by pin holes in YDSE, then we will see a Hyperbolic Fringe pattern. Pointsourcesplacedon theperpendicularaxis tothescreencreateconcentriccircularfringes.
  • 47. Fringe Width 𝐷 𝑂 𝑆1 𝑆2 𝑑 𝝀𝑫 𝒅 Fringe width is the distance between two consecutive maxima/minima. 𝛽 = 3πœ†π· 2𝑑 βˆ’ πœ†π· 2𝑑 𝛽 = πœ†π· 𝑑
  • 48. Fringe Width when the setup is inside a Medium If experimental setup is dipped in liquid πœ‡ = π‘“πœ†π‘£π‘Žπ‘π‘’π‘’π‘š π‘“πœ†π‘šπ‘’π‘‘π‘–π‘’π‘š β‡’ πœ†π‘šπ‘’π‘‘π‘–π‘’π‘š = πœ†π‘£π‘Žπ‘π‘’π‘’π‘š πœ‡ 𝛽 = πœ†π· 𝑑 Fringe Width Refractive index 𝑐 is speed of light in vacuum 𝑣 is speed of light in liquid 𝛽 = πœ†π‘šπ‘’π‘‘π‘–π‘’π‘šπ· 𝑑 β‡’ 𝛽 = πœ†π‘£π‘Žπ‘π‘’π‘’π‘šπ· πœ‡π‘‘ Fringe Width inside liquid, πœ‡ = 𝑐 𝑣
  • 49. Angular Fringe Width 𝑃 𝐡 𝑑 πœƒ 𝑂 𝑦 = 𝛽 𝑆1 𝑆2 𝐷 Central Maxima First Maxima tan πœƒ β‰ˆ πœƒ = 𝛽 𝐷 We know 𝛽 = πœ†π· 𝑑 After putting 𝛽 value in πœƒ = 𝛽 𝐷 πœƒ = πœ† 𝑑
  • 50. Position of Maxima/Minima 𝑦 = π‘›πœ†π· πœ‡π‘‘ 𝑦 = 𝑛 + 1 2 πœ†π· πœ‡π‘‘ ● Constructive Interference: ● Destructive Interference: 𝑦 = Β± πœ†π· πœ‡π‘‘ , Β± 2πœ†π· πœ‡π‘‘ , Β± 3πœ†π· πœ‡π‘‘ . . . . . . . 𝑦 = Β± πœ†π· 2πœ‡π‘‘ , Β± 3πœ†π· 2πœ‡π‘‘ , Β± 5πœ†π· 2πœ‡π‘‘ . . . . . . . 2πœ†π· πœ‡π‘‘ 3πœ†π· 2πœ‡π‘‘
  • 51. Incoherent Light Sources Incoherent Light Sources ● However, the human eye cannot the capture the rapidly changing bright and dark fringes. ● In ultra slow motion, the fringes on the screen flicker. ● So, the eyes see a continuous band of light. ● Net Intensity on the screen is the sum of intensity from two sources.
  • 52. In a Young's double slit interference experiment, the fringe pattern is observed on a screen placed at a distance 𝐷 from the slits. The slits are separated by a distance 𝑑 and are illuminated by monochromatic light of wavelength πœ†. Find the distance from the central point 𝐡 where the intensity falls to half the maximum. 𝐼 = 4𝐼0 cos2 𝛿 2 β‡’ cos 𝛿 2 = 1 2 β‡’ 𝛿 = πœ‹ 2 Phase difference 𝛿 = 2πœ‹ πœ† βˆ†π‘₯ = 𝑦𝑑 𝐷 πœ† 4 = 𝑦𝑑 𝐷 β‡’ β‡’ βˆ†π‘₯ = πœ† 4 𝑦 = πœ†π· 4𝑑 πœ†π· 2𝑑 𝐷 𝑂 𝑆1 𝑆2 𝑑 πΌπ‘šπ‘Žπ‘₯ πΌπ‘šπ‘Žπ‘₯ 2 πœ†π· 4𝑑 𝐡 Path difference βˆ†π‘₯ 4𝐼0 cos2 𝛿 2 = 1 2 4𝐼0 Intensity is half the maximum, Solution: 2πœ‹ πœ† βˆ†π‘₯ = πœ‹ 2
  • 53. Two coherent point sources 𝑆1 and 𝑆2 vibrating in phase emit light of wavelength πœ†. The separation between the sources is 2πœ†. Consider a line passing through 𝑆2 and perpendicular to the line 𝑆1𝑆2. What is the smallest distance from 𝑆2 where the intensity is minimum? T Solution: Given: To find: 𝑑𝑆1𝑆2 = 2πœ† π‘₯ 2πœ† 2 + π‘₯2 βˆ’ π‘₯ = 𝑛 + 1 2 πœ† When π‘₯ > 0, 16πœ† βˆ’ 2𝑛 + 1 2 πœ† > 0 So, 2𝑛 + 1 < 4 𝑛 < 3 2 ∴ 𝑛 = 1 π‘₯ = 16πœ† βˆ’ 2𝑛 + 1 2 πœ† 4 2𝑛 + 1 β‡’ 16πœ† βˆ’ 9πœ† 12 π‘₯ = 7πœ† 12 π‘₯ = 16πœ† βˆ’ 2𝑛 + 1 2 πœ† 4 2𝑛 + 1
  • 54. Solution: T Optical path difference between 𝐡𝑃0 and 𝐴𝑃0 : βˆ†π‘₯ = 𝑑 sin πœƒ β‰ˆ 𝑑 tan πœƒ βˆ†π‘₯ = 𝑑 𝑑 2𝐷 = 𝑑2 2𝐷 βˆ†π‘₯ = πœ† 3 = 𝑑2 2𝐷 𝑑 = 2πœ†π· 3 Figure shows three equidistant slits being illuminated by a monochromatic parallel beam of light. Let 𝐡𝑃0 βˆ’ 𝐴𝑃0 = πœ†/3 and 𝐷 ≫ πœ†. Show that in this case 𝑑 = 2πœ†π·/3.
  • 55. Path Difference between the Two Waves
  • 56. Optical Path length 𝐸 = 𝐸0 sin π‘˜π‘₯0 βˆ’ πœ”π‘‘ Original wave equation: 𝐸 = 𝐸0 sin π‘˜ π‘₯0 + 𝐿 βˆ’ πœ”π‘‘ Equation of a wave when it is ahead by a length 𝐿: 𝐸 = 𝐸0 sin π‘˜π‘₯0 βˆ’ πœ”π‘‘ + π‘˜πΏ Phase difference βˆ†πœ™ π‘˜ = 2πœ‹ πœ† So, βˆ†πœ™ = 2πœ‹ πœ† 𝐿 ∴ Two points on a wave separated by a path length of 𝐿 will have a phase difference of 2πœ‹ πœ† 𝐿. 𝐸 π‘₯ 𝐿 π‘₯0 π‘₯0 + 𝐿
  • 57. Optical Path length 2πœ‹ πœ†π‘Ž πΏπ‘Žπ‘–π‘Ÿ = 2πœ‹ πœ†π‘Ž πœ‡πΏπ‘šπ‘’π‘‘ Phase Difference between points 𝐴 and 𝐡 on the wave, travelling in air: βˆ†πœ™π‘šπ‘’π‘‘ = 2πœ‹ πœ†π‘šπ‘’π‘‘ πΏπ‘šπ‘’π‘‘ Phase Difference between points 𝐴 and 𝐡0 on the wave, travelling in medium: πœ†π‘šπ‘’π‘‘ = πœ†π‘Ž πœ‡ = 2πœ‹ πœ†π‘Ž πœ‡ πΏπ‘šπ‘’π‘‘ βˆ†πœ™π‘šπ‘’π‘‘ = 2πœ‹ πœ†π‘Ž πœ‡πΏπ‘šπ‘’π‘‘ βˆ†πœ™π‘Žπ‘–π‘Ÿ = βˆ†πœ™π‘šπ‘’π‘‘ If, βˆ†πœ™π‘Žπ‘–π‘Ÿ = 2πœ‹ πœ†π‘Ž πΏπ‘Žπ‘–π‘Ÿ πΏπ‘Žπ‘–π‘Ÿ = πœ‡πΏπ‘šπ‘’π‘‘ β‡’ πΏπ‘šπ‘’π‘‘ πœ‡ π‘Žπ‘–π‘Ÿ πΏπ‘Žπ‘–π‘Ÿ 𝐴 𝐡 𝐴 𝐡0 Optical Path Length in a medium is the corresponding path that light travels in vacuum to undergo the same phase difference.
  • 58. 𝑂𝑃𝐷 = 𝑂𝑃𝐿𝐼𝐼 βˆ’ 𝑂𝑃𝐿𝐼 = πœ‡π‘šπ‘’π‘‘ βˆ’ 1 𝐿 Optical Path Difference Optical Path Length in air: 𝑂𝑃𝐿𝐼 = 𝐴𝐡 + πœ‡π‘Žπ‘–π‘ŸπΏ + 𝐢𝐷 = 𝐴𝐡 + 𝐿 + 𝐢𝐷 Optical Path Length in medium: 𝑂𝑃𝐿𝐼𝐼 = 𝐴𝐡 + πœ‡π‘šπ‘’π‘‘πΏ + 𝐢𝐷 Optical Path Difference: Phase Difference: 𝛿 = βˆ†πœ™πΌπΌ βˆ’ βˆ†πœ™πΌ = 2πœ‹ πœ†π‘Ž πœ‡π‘šπ‘’π‘‘ βˆ’ 1 𝐿 𝛿 = 2πœ‹ πœ†π‘Ž 𝑂𝑃𝐷 𝑂𝑃𝐷 = 𝑂𝑃𝐿𝐼𝐼 βˆ’ 𝑂𝑃𝐿𝐼 = πœ‡2 βˆ’ πœ‡1 𝐿 and Phase Difference: 𝛿 = βˆ†πœ™πΌπΌ βˆ’ βˆ†πœ™πΌ βˆ†πœ™πΌπΌ = 2πœ‹ πœ†π‘Ž πœ‡2𝐿 𝛿 = 2πœ‹ πœ†π‘Ž πœ‡2 βˆ’ πœ‡1 𝐿 Phase Difference in medium 1 and 2: Optical Path Difference: βˆ†πœ™πΌ = 2πœ‹ πœ†π‘Ž πœ‡1𝐿
  • 59. Thin Transparent Film in YDSE Optical Path Difference: 𝑂𝑃𝐷 = 𝐼 βˆ’ 𝐼𝐼 = πœ‡π‘‘ + 𝑆1𝐡 βˆ’ 𝑑 βˆ’ 𝑆2𝐡 Optical Path Travelled by wave 𝐼: πœ‡π‘‘ + 𝑆1𝐡 βˆ’ 𝑑 𝑆2𝐡 Optical Path Travelled by wave 𝐼𝐼: 𝑆1𝐡 = 𝑆2𝐡 𝑂𝑃𝐷 = πœ‡π‘‘ βˆ’ 𝑑 = πœ‡ βˆ’ 1 𝑑 β€’ Introduction of slab causes change in 𝑂𝑃𝐷 by πœ‡ βˆ’ 1 𝑑. β€’ Path length of the ray 𝑆1𝐡 increases when it encounters the thin film. β€’ Now the central maxima does not lie at 𝐡.
  • 60. Thin Transparent Film in YDSE Optical Path Difference: βˆ†π‘₯ = 𝑆2𝑃 βˆ’ 𝑆1𝑃 βˆ†π‘₯ = 𝑦𝑑 𝐷 βˆ†π‘₯ = 𝐼𝐼 βˆ’ 𝐼 Total Optical Path Difference at point P: βˆ†π‘₯ = 𝐼𝐼 βˆ’ 𝐼 βˆ†π‘₯ = 𝑦𝑑 𝐷 βˆ’ 𝑑 πœ‡ βˆ’ 1 β€’ After inserting thin film, light ray 𝑆1𝑃 has travelled an extra path of πœ‡ βˆ’ 1 𝑑. βˆ†π‘₯ = 𝑆2𝑃 βˆ’ 𝑆1𝑃 + 𝑑 πœ‡ βˆ’ 1 β€’ After inserting thin film, light ray 𝑆2𝑃 has travelled an extra path of πœ‡ βˆ’ 1 𝑑. Total Optical Path Difference at point P: βˆ†π‘₯ = 𝐼𝐼 βˆ’ 𝐼 βˆ†π‘₯ = 𝑦𝑑 𝐷 + 𝑑 πœ‡ βˆ’ 1 βˆ†π‘₯ = 𝑆2𝑃 + 𝑑 πœ‡ βˆ’ 1 βˆ’ 𝑆1𝑃
  • 61. Shift in Central Maxima 𝑃 𝐡 𝑑 πœƒ 𝑂 𝑦 π‘₯ π‘₯ + Ξ”π‘₯ 𝐷 𝑑, πœ‡ πœ‡ βˆ’ 1 𝑑𝐷 𝑑 At central maxima, 𝑆1 𝑆2 𝐼 𝐼𝐼 Screen 𝑂𝑃𝐷 = 0 𝑦𝑑 𝐷 βˆ’ πœ‡ βˆ’ 1 𝑑 = 0 𝑦 = πœ‡ βˆ’ 1 𝑑𝐷 𝑑 Shift in central maxima =
  • 62. Number of Fringes Shifted πœ‡ βˆ’ 1 𝑑𝐷 𝑑 Shift in central maxima: Number of Fringes shifted 𝑛 = 𝑦 𝛽 = πœ‡ βˆ’ 1 𝑑𝐷 𝑑 Γ— 𝑑 πœ†π· 𝑛 = πœ‡ βˆ’ 1 𝑑 πœ† = π‘ β„Žπ‘–π‘“π‘‘ 𝑖𝑛 π‘π‘’π‘›π‘‘π‘Ÿπ‘Žπ‘™ π‘šπ‘Žπ‘₯π‘–π‘šπ‘Ž π‘“π‘Ÿπ‘–π‘›π‘”π‘’ π‘€π‘–π‘‘π‘‘β„Ž 𝑆 = πœ‡ βˆ’ 1 𝑑𝐷 𝑑 Shift in central maxima: Number of fringes shifted: 𝑛 = πœ‡ βˆ’ 1 𝑑 πœ†
  • 63. T The Young’s double slit experiment is done in a medium of refractive index 4/3. A light of 600 π‘›π‘š wavelength is falling on the slits having 0.45 π‘šπ‘š separation. The lower slit 𝑆2 is covered by a thin glass sheet of thickness 10.2 πœ‡π‘š and refractive index 1.5. The interference pattern is observed on a screen placed 1.5 π‘š from the slits. Find the location of the central maximum on the 𝑦 βˆ’axis. 10.2π‘šπ‘š 7.5 π‘šπ‘š 1.25 π‘šπ‘š 4.25 π‘šπ‘š B A D C 𝑃0 𝑂 πœ‡ = 4/3 𝑆1 𝑆2 Screen 𝐼 𝐼𝐼
  • 64. Given: Solution: To find: πœ‡π‘šπ‘’π‘‘ = 4 3 , 𝑑 = 0.45 π‘šπ‘š, 𝐷 = 1.5 π‘š, πœ† = 600 π‘›π‘š, 𝑑 = 10.2 πœ‡π‘š, πœ‡π‘‘ = 1.5 Position of central maxima 𝑂𝑃𝐷 = 𝐼 βˆ’ 𝐼𝐼 𝑂𝑃𝐷 = 𝑆1𝑃0 πœ‡π‘šπ‘’π‘‘ βˆ’ πœ‡π‘”π‘‘ + 𝑆2𝑃0 πœ‡π‘šπ‘’π‘‘ βˆ’ πœ‡π‘šπ‘’π‘‘π‘‘ 𝑂𝑃𝐷 = 𝑆1𝑃0 βˆ’ 𝑆2𝑃0 πœ‡π‘šπ‘’π‘‘ βˆ’ 𝑑 πœ‡π‘” βˆ’ πœ‡π‘šπ‘’π‘‘ 𝑂𝑃𝐷 = 𝑦𝑑 𝐷 πœ‡π‘šπ‘’π‘‘ βˆ’ 𝑑 πœ‡π‘” βˆ’ πœ‡π‘šπ‘’π‘‘ At central maxima 𝑂𝑃𝐷 = 0 𝑦𝑑 𝐷 πœ‡π‘šπ‘’π‘‘ βˆ’ 𝑑 πœ‡π‘” βˆ’ πœ‡π‘šπ‘’π‘‘ = 0 𝑦 = 𝑑𝐷 π‘‘πœ‡π‘šπ‘’π‘‘ πœ‡π‘” βˆ’ πœ‡π‘šπ‘’π‘‘ Substituting all the values we get, 𝑦 = 4.25 π‘šπ‘š 𝑂 πœ‡ = 4/3 𝑆1 𝑆2 Screen 𝑃0 𝐼 𝐼𝐼 𝑦
  • 65. Given: Solution: To find: πœ† = 5000 ሢ 𝐴, πœ‡ = 1.5, 𝑑 𝑦 = πœ‡ βˆ’ 1 𝑑𝐷 𝑑 𝑦5𝐡 = 5π·πœ† 𝑑 𝑦𝑛𝐡 = π‘›π·πœ† 𝑑 Shift in central maxima: πœ‡ βˆ’ 1 𝑑𝐷 𝑑 = 5π·πœ† 𝑑 𝑑 = 5πœ† πœ‡ βˆ’ 1 = 5 Γ— 5000 Γ— 10βˆ’10 1.5 βˆ’ 1 𝑑 = 5 πœ‡π‘š T In YDSE, find the thickness of a glass slab (πœ‡ = 1.5) which should be kept in front of upper slit 𝑆1 so that the central maxima is formed at a place where 5π‘‘β„Ž bright fringe was lying earlier (before inserting slab). πœ† = 5000 ሢ 𝐴
  • 66. Given: Solution: To find: Wavelength = πœ†, 𝑑1 = 𝑑2 = 𝑑 =? Path difference due to both the slabs at 𝑃0: For 𝑑 to be minimum, put 𝑛 = 0, β‡’ πœ‡1 βˆ’ πœ‡2 𝑑 = 𝑛 + 1 2 πœ† Ξ”π‘₯ = 𝑛 + 1 2 πœ† For minima at 𝑃0, we know, πœ‡1 βˆ’ 1 𝑑 βˆ’ πœ‡2 βˆ’ 1 𝑑 = πœ‡1 βˆ’ πœ‡2 𝑑 Refractive index = πœ‡1 & πœ‡2 Screen 𝐷 𝑃0 𝑑 𝑂 𝑆1 𝑆2 𝑑 πœ‡1 πœ‡2 T 𝑑 = πœ† 2 πœ‡1 βˆ’ πœ‡2 Two transparent slabs, having equal thickness but different refractive indices πœ‡1 and πœ‡2 πœ‡1 > πœ‡2 , are pasted side by side to form a composite slab. This slab is placed just after the double slits in a Young’s experiment so that the light from one slit goes through one material and light through other slit goes through other material. What should be the minimum thickness of the slab so that there is a minimum at point 𝑃0 which is equidistant from the slits?
  • 67. Phase change Yes No Reflected from Denser Medium Rarer Medium 𝑖 π‘Ÿ Air Air πœ‡ No phase change No phase change πœ‹ Phase change in Reflection & Refraction
  • 68. A narrow slit 𝑆 transmitting light of wavelength πœ† is placed at a distance 𝑑 above a large plane mirror. The light coming directly from the slit and that coming after reflection interfere at a screen placed at a distance 𝐷 from the slit. (π‘Ž) What will be the intensity at a point just above 𝑂. Given: To find: Solution: Wavelength = πœ†, Distance of screen = 𝐷, 𝑆𝑃 β‰ˆ 𝑆𝑄𝑃 β†’ path difference = 0 (when 𝑃 is just above 𝑂) Screen 𝑂 𝐷 𝑆 𝑑 𝑄 𝑃 Ξ΄ = πœ‹ (due to reflection of light from a denser medium) Distance of source from the mirror = 𝑑 Intensity just above 𝑂 T 𝐼𝑛𝑒𝑑 = 0 If phase difference is odd integral multiple of πœ‹, destructive interference will take place.
  • 69. A narrow slit 𝑆 transmitting light of wavelength πœ† is placed at a distance 𝑑 above a large plane mirror. The light coming directly from the slit and that coming after reflection interfere at a screen placed at a distance 𝐷 from the slit. Given: To find: Solution: T (𝑏) At what distance from 𝑂 does the first maximum occur? Screen 𝑂 𝐷 𝑆1 𝑑 𝑄 𝑃 𝑆2 𝑑 πœƒ 𝑦 Wavelength = πœ†, 𝑦1𝑠𝑑 π‘šπ‘Žπ‘₯π‘–π‘šπ‘Ž Distance of screen = 𝐷, Distance of source from the mirror = 𝑑 βˆ†π‘₯ = 2𝑑 sin πœƒ = 2𝑑 tan πœƒ = 2𝑑𝑦 𝐷 βˆ†π‘₯𝑛𝑒𝑑 = 2𝑑𝑦 𝐷 + πœ† 2 = π‘›πœ† = 1 πœ† β‡’ 2𝑑𝑦 𝐷 = πœ† 2 𝑦 = πœ†π· 4𝑑
  • 70. Given: To find: Solution: πœ† = 700 π‘›π‘š, 𝐷 = 1 π‘š, 𝛽 = 2 Γ— πœ†π· 4𝑑 = πœ†π· 2𝑑 𝛽 = 700 Γ— 10βˆ’9 Γ— 1 2 Γ— 10βˆ’3 = 3.5 Γ— 10βˆ’4 π‘š 𝑑 = 1 π‘šπ‘š Fringe width = 𝛽 T A narrow slit 𝑆 transmitting light of wavelength πœ† is placed at a distance 𝑑 above a large plane mirror. The light coming directly from the slit and that coming after reflection interfere at a screen placed at a distance 𝐷 from the slit. (𝑐) If 𝑑 = 1 π‘šπ‘š, 𝐷 = 1 π‘š and πœ† = 700 π‘›π‘š, then find the fringe width. 𝑂 𝐷 𝑆1 𝑑 𝑄 𝑃 𝑆2 𝑑 πœƒ πœ†π· 4𝑑 πœ†π· 4𝑑 𝛽 = 0.35 π‘šπ‘š
  • 71. A narrow slit 𝑆 transmitting light of wavelength πœ† is placed at a distance 𝑑 above a large plane mirror. The light coming directly from the slit and that coming after reflection interfere at a screen placed at a distance 𝐷 from the slit. (𝑑) If the mirror reflects only 64 % of the light energy falling on it, what will be the ratio of the maximum to the minimum intensity in the interference pattern observed on the screen? Given: To find: Solution: πΌπ‘šπ‘Žπ‘₯ = 𝐼1 + 𝐼2 2 , πΌπ‘šπ‘Žπ‘₯ πΌπ‘šπ‘–π‘› T πΌπ‘šπ‘Žπ‘₯ πΌπ‘šπ‘–π‘› = 81 𝑂 𝐷 𝑆1 𝑄 𝑃 𝑆2 πœƒ 𝐼1 = 𝐼0 𝐼2 = 0.64 𝐼0 Wavelength = πœ†, Distance of screen = 𝐷, Distance of source from the mirror = 𝑑 𝐼2 = 0.64 𝐼0 πΌπ‘šπ‘–π‘› = 𝐼1 βˆ’ 𝐼2 2 πΌπ‘šπ‘Žπ‘₯ πΌπ‘šπ‘–π‘› = 𝐼1 + 𝐼2 2 𝐼1 βˆ’ 𝐼2 2 β‡’ 𝐼0 + 0.64 𝐼0 2 𝐼0 βˆ’ 0.64 𝐼0 2 = 1 + 0.8 2 1 βˆ’ 0.8 2 = 1.8 2 0.2 2
  • 72. Condition for Thin Film Interference 𝑑 𝑖 π‘Ÿ air air πœ‡ πœ‹ β€’ The film thickness should be comparable to the wavelength of light 𝑑 β‰ˆ πœ† . β€’ The incident light should be white (non - monochromatic). β€’ The angle of incidence should be small 𝑖 β‰ˆ 0.
  • 73. Thin Film Interference β€’ Inside the film, when a particular colour’s path difference is even integral multiple of the wavelength, it undergoes constructive interference, so these colours appear bright. β€’ Interference of light wave being reflected off two surfaces that are at a distance comparable to its wavelength is known as thin film ”Thin film interference”. β€’ When a particular colour’s wavelength is odd integral multiple of the path difference inside the film, it undergoes destructive interference, so these colours don’t appear at all. 𝑖 π‘Ÿ Air Air πœ‡ No phase change No phase change πœ‹
  • 74. βˆ†π‘₯ = 2πœ‡π‘‘ Interference due to Thin Film from Transmitted Light angles are very small 𝑑 𝑖 π‘Ÿ air air πœ‡ Path difference of 1𝑠𝑑 and 2𝑛𝑑 transmitted light wave π‘Ÿ 𝑑 𝑑 cos π‘Ÿ = 2 𝑑 cos π‘Ÿ πœ‡ When angle is very small, π‘Ÿ β‰ˆ 0 cos π‘Ÿ β‰ˆ 1 Path difference π‘Ÿ 1𝑠𝑑 2𝑛𝑑
  • 75. For constructive interference: Or For destructive interference: Or 2πœ‡π‘‘ = π‘›πœ† 2πœ‡π‘‘ = 𝑛 + 1 2 πœ† 𝛿 = 2π‘›πœ‹ 𝛿 = 2𝑛 + 1 πœ‹ angles are very small 𝑑 𝑖 π‘Ÿ air air πœ‡ Interference due to Thin Film from Transmitted Light
  • 76. βˆ†π‘₯ = 2πœ‡π‘‘ βˆ’ πœ† 2 Path difference of 1𝑠𝑑 and 2𝑛𝑑 reflected light wave π‘Ÿ 𝑑 𝑑 cos π‘Ÿ = 2 𝑑 cos π‘Ÿ πœ‡ β‰ˆ 2πœ‡π‘‘ Total path difference 𝑑 𝑖 π‘Ÿ air air πœ‡ angles are very small 1𝑠𝑑 2𝑛𝑑 Interference due to Thin Film from Reflected Light Because phase change of πœ‹ after reflection of 1𝑠𝑑 , path difference = βˆ’ πœ† 2
  • 77. For constructive interference: For destructive interference: 2πœ‡π‘‘ = π‘›πœ† 2πœ‡π‘‘ = 𝑛 + 1 2 πœ† 𝑑 𝑖 π‘Ÿ air air πœ‡ πœ‹ angles are very small Interference due to Thin Film from Reflected Light 2πœ‡π‘‘ βˆ’ πœ† 2 = π‘›πœ† 2πœ‡π‘‘ βˆ’ πœ† 2 = 𝑛 + 1 2 πœ† β‡’ 2πœ‡π‘‘ = 𝑛 + 1 πœ† 𝑛 is an integer.
  • 78. Interference due to Thin Film For constructive interference: β€’ Colours will be strongly reflected/transmitted. Destructive interference: β€’ Colours will be poorly reflected/transmitted. This gives coloured appearance of the film.
  • 79. A soap film of refractive index 1.33 is illuminated by the light of wavelength 400 π‘›π‘š at an angle of 45Β°. If there is complete destructive interference then, find the thickness of the film. Given: To find: Solution: T πœ‡ = 1.33, 𝑖 = 45Β°, πœ† = 400 π‘›π‘š 𝑑 πœ‡ = sin 𝑖 sin π‘Ÿ β‡’ 1.33 = sin 45 sin π‘Ÿ sin π‘Ÿ = 3 4 2 cos π‘Ÿ = 1 βˆ’ sin2 π‘Ÿ For Destructive interference: 2πœ‡π‘‘ cos π‘Ÿ = π‘›πœ† β‡’ 2 1.33 𝑑 0.85 = 1 400 Γ— 10βˆ’9 β‡’ 𝑑 = 1.27 Γ— 10βˆ’7 π‘š β‡’ cos π‘Ÿ = 0.85 45Β° soap film
  • 80. Diffraction β€’ Bending of a wave or its deviation from its original direction of propagation while passing through a small obstruction is known as diffraction. Original direction β€’ Diffraction is explained by wave nature of light.​ β€’ Condition for bending: size of obstacle β‰ˆ πœ† β€’ Every point on the wavefront makes a secondary wave according to Huygens Principle.
  • 81. β€’ Source and screen are at infinite distance from diffraction element. β€’ Source and screen are at finite distance from diffraction element. β€’ High intense interference pattern is observed on screen. β€’ Diffraction was first demonstrated by this experiment. 𝑆𝑂 𝑆𝑂 Diffraction of Light Waves Fraunhofer Diffraction Fresnel Diffraction
  • 82. 𝑃𝑂 𝑏 𝑃 πœƒ Screen 𝑏 2 𝑏 2 sin πœƒ πœƒ β€’ Path difference between the waves, βˆ†x = 𝑏 2 sin πœƒ Fraunhofer Diffraction – Path Difference
  • 83. 𝑃𝑂 𝑏 𝑃 πœƒ Screen β€’ The condition for 1𝑠𝑑 dark fringe formed at point 𝑃: Fraunhofer Diffraction – first Dark Fringe 𝑏 2 sin πœƒ = πœ† 2 𝑏 sin πœƒ = πœ† β€’ The condition for π‘›π‘‘β„Ž dark fringe: 𝑏 sin πœƒ = π‘›πœ† β€’ Central Maxima is at 𝑃0, where πœƒ = 0
  • 84. 𝑃𝑂 𝑏 𝑃 πœƒ Screen β€’ Electric Field Amplitude at 𝑃: Fraunhofer Diffraction – Intensity at general point 𝐸′ = 𝐸0 sin 𝛽 𝛽 β€’ When πœƒ = 0, 𝛽 = 0, 𝐸′ = 𝐸0 𝐼 = 𝐼𝑂 sin2 𝛽 𝛽2 Where 𝐼0 is the intensity at Central Maxima Where 𝛽 = 2πœ‹ πœ† 𝑏 2 sin πœƒ 𝛽 = phase difference β€’ Intensity ∝ 𝐸2 β€’ Intensity at a general point:
  • 85. Fraunhofer Diffraction – Graph of Intensity 𝐼 = 𝐼𝑂 sin2 𝛽 𝛽2 𝛽 = πœ‹π‘ sin πœƒ πœ† πΌπ‘œ βˆ’ 2πœ† 𝑏 sin πœƒ 𝐼 0 βˆ’ πœ† 𝑏 πœ† 𝑏 2πœ† 𝑏 𝐼𝑂 22 𝐼𝑂 62.5 β€’ sin πœƒ = 0 ; Central Maxima β€’ sin πœƒ = ; Minima π‘›πœ† 𝑏 β€’ sin πœƒ = Β± πœ† 𝑏 , Β± 2πœ† 𝑏 , Β± 3πœ† 𝑏 … … … … β€’ Maximum intensity is distributed between βˆ’ πœ† 𝑏 and πœ† 𝑏 . β€’ If 𝑏 is large, πœ† 𝑏 β†’ 0, then Single fringe is observed and no diffraction is seen.
  • 86. Fraunhofer Diffraction – Graph of Intensity 𝑏 β€’ Fraunhofer Diffraction ⟢ Intensity decreases away from the centre. β€’ 𝐼0 β†’ Intensity at the central bright fringe β€’ YDSE ⟢ Intensity is same for all fringes. β€’ 𝐼0 β†’ Intensity from a single slit. 𝐼 = 𝐼𝑂 sin2 𝛽 𝛽2 𝐼 = 4𝐼𝑂 cos2 πœ™ 2
  • 87. Diffraction Difference between Interference and Diffraction Interference It is the phenomenon of superposition of two waves coming from two different coherent sources. It is the phenomenon of superposition of two waves coming from two different parts of the same wavefront. In interference pattern, all bright fringes are equally bright and equally spaced. All bright fringes are not equally bright and equally wide. Brightness and width decreases with the angle of diffraction. All dark fringes are perfectly dark. All dark fringes are perfectly dark, but their contrast with bright fringes and width decreases with angle of diffraction. In interference, bright and dark fringes are large in number for a given field of view. In diffraction, bright and dark fringes are fewer for a given field of view.
  • 88. A beam of light of wavelength 600 π‘›π‘š from a distant source falls on a single slit 1 π‘šπ‘š wide and a resulting diffraction pattern is observed on a screen 2 π‘š away. The distance between the first dark fringes on either side of central bright fringe is 1π‘šπ‘š 𝛼 πœƒ 𝐷 = 2 π‘š πœƒ sin πœƒ β‰ˆ πœƒ Solution: πœƒ = πœ† 𝑏 𝛼 = 2πœƒ = 2πœ† 𝑏 𝛼 = 𝑦 𝐷 = 2πœ† 𝑏 ⟹ 𝑦 = 2πœ†π· 𝑏 𝑦 = 2 Γ— 600 Γ— 10βˆ’9 Γ— 2 10βˆ’3 = 24 Γ— 10βˆ’4 π‘š 𝑦 = 2.4 π‘šπ‘š First maxima is formed at πœ† 𝑏 distance away on both side of central maxima. 𝑦 (πœƒ is small)
  • 89. 𝐴 𝐡 𝑁 3πœ† 2 𝑏 𝑏 sin πœƒ = 3πœ† 2 πœƒ1𝐡 = 3πœ† 2𝑏 𝑦1𝐡 = 3π·πœ† 2𝑏 β€’ Condition for 1𝑠𝑑 secondary maxima (bright): β€’ The angle of diffraction (πœƒ1B) for 1st maximum (bright) is: β€’ The position of 1st maximum (bright) from the centre of the screen is: Single Slit Diffraction – πŸπ’”π’• Secondary Maxima
  • 90. Screen First secondary maxima Second secondary maxima Central maxima First secondary maxima Second secondary maxima Where, 𝑛 = 1, 2, 3, 4 … … … β€’ The angle of diffraction πœƒπ‘› for π‘›π‘‘β„Ž maxima (bright) is: β€’ The position of π‘›π‘‘β„Ž maxima (bright) from the centre of the screen is: πœƒπ‘›π΅ = 2𝑛 + 1 πœ† 2𝑏 𝑦𝑛𝐡 = 2𝑛 + 1 π·πœ† 2𝑏 Single Slit Diffraction – 𝒏𝒕𝒉 Secondary Maxima
  • 91. β€’ Angular position for minima + πœ† 𝑏 βˆ’ πœ† 𝑏 + 2πœ† 𝑏 βˆ’ 2πœ† 𝑏 β€’ Angular position for maxima sin πœƒ = Β± πœ† 𝑏 , Β± 2πœ† 𝑏 , … … , Β± π‘›πœ† 𝑏 sin πœƒ = Β± 3πœ† 2𝑏 , Β± 5πœ† 2𝑏 , … … , Β± (2𝑛 + 1)πœ† 2𝑏 Single Slit Diffraction + 3πœ† 2𝑏 βˆ’ 3πœ† 2𝑏 + 5πœ† 2𝑏 βˆ’ 5πœ† 2𝑏 0
  • 92. Given: To find: Solution: Angular position of first minima is given by, sin πœƒ1𝐷 = πœ† π‘Ž ⟹ 1 2 = πœ† π‘Ž Angular position of first secondary maxima is given by, sin πœƒ1𝐡 = 3πœ† 2π‘Ž ⟹ sin πœƒ1𝐡 = 3 4 πœƒ1𝐡 = sinβˆ’1 3 4 T πœƒ1𝐷 = 30Β°; πœ† = 5000 ሢ 𝐴 Angular position of first secondary maximum (πœƒ1𝐡) In a diffraction pattern due to a single slit of width π‘Ž, the first minima is observed at an angle 30Β° when the light of wavelength 5000 ሢ 𝐴 is incident on the slit. The first secondary maxima is observed at an angle of:
  • 93. β€’ When a monochromatic light is incident on the hole we see concentric circular bright and dark spots on the screen. β€’ The size of hole should be comparable to the wavelength of incident light. β€’ Central bright spot contains the most energy. β€’ Brightness of the rings decreases as we move away from the centre. β€’ For the first dark ring, sin πœƒ = 1.22πœ† 𝑏 Fraunhofer Diffraction for Hole 𝑏 𝐷 πœƒ 𝑅 𝑅 = 1.22πœ†π· 𝑏
  • 94. A convex lens of diameter 8.0 π‘π‘š is used to focus a parallel beam of light of wavelength 6200 Γ…. If the light be focused at a distance of 20 π‘π‘š from the lens, what would be the radius of the central bright spot? 6200 Γ… 𝑏 = 8 π‘π‘š πœƒ 20 π‘π‘š Given: To Find: Solution: 𝐷 = 20 π‘π‘š, 𝑏 = 8 π‘π‘š and πœ† = 6200 Γ… 𝑅 Radius of central bright spot, 𝑅 = 1.22πœ†π· 𝑏 β‡’ 𝑅 = 1.22 Γ— 6200 Γ— 10βˆ’10 Γ— 20 Γ— 10βˆ’2 8 Γ— 10βˆ’2 𝑅 = 1.89 Γ— 10βˆ’6 π‘š 𝑅
  • 96. Limit of Resolution Just resolved: Well resolved: Unresolved: Diffraction discs from both sources overlap. 𝑆1 𝑆2 𝑆1 𝑆2 𝑆1 𝑆2 The periphery of the diffraction disc of one object touches the centre of the diffraction disc of the other object. The diffraction discs formed by two objects are well separated from each other.
  • 97. Rayleigh criterion The Rayleigh criterion specifies the minimum separation between two light sources that may be resolved into distinct objects. Unresolved Just resolved Well resolved (Rayleigh criterion)
  • 98. Limit of Resolution of a Telescope β€’ Clear image is formed when the diffraction discs from two sources is just resolved. π‘†π‘‘π‘Žπ‘Ÿ1 π‘†π‘‘π‘Žπ‘Ÿ2 Just resolved 𝑅 βˆ†πœƒ Telescope lens β€’ Distance between diffraction disc > 𝑅 β‡’ Well resolved β€’ Distance between diffraction disc < 𝑅 β‡’ Unresolved
  • 99. Resolving Power of Telescope β€’ Angular limit of resolution of telescope: βˆ†πœƒ = 1.22πœ† 𝑏 βˆ†πœƒ = Limit of resolution β€’ Angle subtended by the first dark fringe > βˆ†πœƒ β‡’ Well resolved β€’ Angle subtended by the first dark fringe < βˆ†πœƒ β‡’ Unresolved 𝑏 𝑓 𝑅 βˆ†πœƒ
  • 100. Radius of the Central Bright Spot β€’ Radius of central bright region is: β€’ Resolving power of a telescope: 𝑅 = 1.22πœ†π‘“ 𝑏 𝑅. 𝑃. = 𝑏 1.22πœ† 𝑅 = π‘“βˆ†πœƒ 𝑅. 𝑃. = 1 βˆ†πœƒ β€’ Bigger lens β‡’ larger 𝑏 β‡’ smaller Limit of Resolution βˆ†πœƒ β‡’ higher Resolving Power 𝑅. 𝑃. 𝑏 𝑓 𝑅 βˆ†πœƒ
  • 101. Disadvantages of using Lens 1. Difficult and expensive to build large lenses. 2.Providing mechanical support to large lenses require large and complex machinery. 3. Chromatic aberration ( light rays passing through a lens focus at different points, depending on their wavelength). The largest lens objective in use has diameter of 40 inch (~1.02 π‘š). It is at the Yerkes Observatory in Wisconsin, USA. Fact: Chromatic aberration
  • 102. Advantages of using Mirror as the Objective in Telescope 1. No chromatic aberration 2. Parabolic mirror used to counter spherical aberration. 3. Large mirrors can be supported from the back. The viewer sits near the focal point of the mirror, in a small cage. Fact: Eyepiece Objective mirror
  • 103. Given: To find: Solution: Limit of resolution of telescope is given by: βˆ†πœƒ = 1.22 Γ— 500 Γ— 10βˆ’9 200 Γ— 10βˆ’2 T 𝑏 = 200 π‘π‘š, πœ† = 500 π‘›π‘š Limit of resolution of telescope 305 Γ— 10–9 π‘Ÿπ‘Žπ‘‘π‘–π‘Žπ‘› βˆ†πœƒ = 1.22πœ† 𝑏 Calculate the limit of resolution of a telescope objective having a diameter of 200 π‘π‘š, if it has to detect light of wavelength 500 π‘›π‘š coming from a star. JEE Main 2019
  • 104. Limit of Resolution of a Microscope Angular limit of resolution of microscope: sin πœƒ β‰ˆ πœƒ = 1.22πœ† 𝑏 𝑅 = π‘£πœƒ = 𝑣 Γ— 1.22πœ† 𝑏 π‘š = 𝑅 𝑑 β‡’ 𝑑 = 𝑅 π‘š Magnification of convex lens: 𝑑 = 𝑣 Γ— 1.22πœ† π‘π‘š β€’ Clear images can be seen in the microscope if the diffraction discs are just resolved. Lens Formula: 1 𝑣 βˆ’ 1 𝑒 = 1 𝑓 1 βˆ’ 𝑣 𝑒 = 𝑣 𝑓 π‘š = 1 βˆ’ 𝑣 𝑓 β‡’ 1 βˆ’ π‘š = 𝑣 𝑓 π‘š = 𝑣 𝑒 π‘š β‰ˆ βˆ’ 𝑣 𝑓 𝑣 ≫ 𝑓 β‡’ 𝑣 𝑓 ≫ 1 𝑑 = 𝑣 Γ— 1.22πœ† π‘π‘š We know that: 𝑑 = 1.22πœ† π‘π‘š Γ— βˆ’π‘šπ‘“ = 1.22πœ†π‘“ 𝑏 tan 𝛽 = 𝑏 2𝑓 β‡’ 𝑏 = 2𝑓 tan 𝛽 𝛽 is small, 𝑏 = 2𝑓 sin 𝛽 π‘‘π‘šπ‘–π‘› = 1.22πœ†π‘“ 𝑏 = 1.22πœ†π‘“ 2𝑓 sin 𝛽 π‘‘π‘šπ‘–π‘› = 1.22πœ† 2 sin 𝛽
  • 105. Microscope immersed in Oil When the setup is immersed in oil, πœ†π‘šπ‘’π‘‘ changes to πœ† πœ‡ π‘‘π‘šπ‘–π‘› = 1.22πœ†π‘šπ‘’π‘‘ 2 sin 𝛽 ⟢ π‘‘π‘šπ‘–π‘› = 1.22πœ† 2πœ‡ sin 𝛽 Note: The product πœ‡sin𝛽 is called the numerical aperture and is sometimesmarked on the objective. Resolving power of a microscope: β€’ Resolving power of the microscope increases when it is immersed in a medium. β‡’ 𝑅. 𝑃. = 2πœ‡ sin 𝛽 1.22πœ† 𝑅. 𝑃. = 1 π‘‘π‘šπ‘–π‘›
  • 106. Validity of Ray Optics β€’ Consider a single slit of width β€˜π‘Žβ€™.Diffraction pattern will be observed. β€’ There will be central maxima due to diffraction. β€’ Angular size of central maximum, πœƒ = πœ† π‘Ž If 𝑦 β‰ˆ π‘Ž β€’ The width of diffracted beam after it has travelled by 𝑧, 𝑦 = 𝑧 Γ— πœƒ = πœ†π‘§ π‘Ž 𝑧𝐹 is Fresnel distance. β€’ If 𝑧 < 𝑧𝐹, the ray optics is valid. β€’ If 𝑧 > 𝑧𝐹, spreading due to diffraction dominates. 𝑧 β‰ˆ π‘Ž2 πœ† = 𝑧𝐹
  • 107. Given: 𝑧𝐹 = 2 Γ— 10βˆ’3 2 600 Γ— 10βˆ’9 = 6.7 π‘š T π‘Ž = 2 π‘šπ‘š; πœ† = 600 π‘›π‘š Solution: Ray optics is a good approximation up to Fresnel distance only. 𝑧 β‰ˆ π‘Ž2 πœ† = 𝑧𝐹 β€’ If 𝑑 < 6.7 π‘š ⟢ Ray optics holds. β€’ If 𝑑 > 6.7 π‘š ⟢ Spreading due to diffraction dominates. For what distance is ray optics a good approximation when a plane light wave is incident on a circular aperture of width 2 π‘šπ‘š having wavelength 600 π‘›π‘š?
  • 108. When a parallel beam of light passes through a medium, a part of it appears in directions other than the incident direction. This phenomenon is called scattering of light. Aparticle Scattering β€’ Light is an EM wave, it oscillates the charged particles in a medium because of its oscillating electric field. β€’ Oscillating charged particles emit EM waves. β€’ If the oscillating electric field of incident light has frequency 𝑓, the frequency of scattered wave will also have frequency 𝑓. Incidentlight
  • 109. Rayleigh’s Law of Scattering β€’ When size of particles < πœ† β€’ Intensity of scattering depends on Intensity of scattered wave ∝ 1 πœ†4 1. Wavelength of light 2. Size of particles causing scattering 𝑉 𝐼 𝐡 π‘Œ 𝑂 𝑅 𝐺 πœ† βˆ’ π‘šπ‘–π‘›π‘–π‘šπ‘’π‘š πœ† βˆ’ π‘šπ‘Žπ‘₯π‘–π‘šπ‘’π‘š Scatters most Scatters least πœ† increases and Intensity of scattering decreases
  • 110. Unpolarized Light ● The light having electric field oscillations in all directions in the plane perpendicular to the direction of propagation. Note: Light wave is coming out of the screen, and arrows show direction of oscillation of electric field. ● Examples of source of unpolarized light: Candle, Bulb, Sun etc.
  • 111. Plane Polarized Light Plane Polarized light – When electric field at a point always remains parallel to a fixed direction as time passes. Plane of polarization – Plane containing electric field and direction of propagation. π‘₯ βˆ’ π‘Žπ‘₯𝑖𝑠 β†’ Direction of propagation. 𝑦 βˆ’ π‘Žπ‘₯𝑖𝑠 β†’ Oscillation of electric field. π‘₯𝑦 π‘π‘™π‘Žπ‘›π‘’ β†’ Plane of polarization. When the polarizer is placed in the path of unpolarized light, the direction of oscillation of the electric field becomes parallel to the transmission axis. Note: If any unpolarized light of intensity 𝐼0 is incident on a polarizer, we get a polarized light of intensity . 𝐼0 2
  • 112. Law of Malus ● 𝐼 = Intensity of transmitted light ● 𝐼0 = Intensity of incident light Intensity of the wave after crossing the polarizer: Electric field amplitude of the wave after crossing the polarizer: 𝐸 = 𝐸0 cos πœƒ 𝐼 ∝ 𝐸2 𝐼 = 𝐼0 cos2 πœƒ πœƒ 𝐸0 𝐸0 cos πœƒ 𝐼0 𝐼 Polarized light Unpolarized light 2𝐼0
  • 113. Law of Malus 𝐼 = 𝐼0 cos2 0Β° = 𝐼0 ● Case 1 β†’ when πœƒ = 0Β° : ● Case 2 β†’ when πœƒ = 90Β° : 𝐼 = 𝐼0 cos2 90Β° = 0 𝐼0 𝐼0 𝐼0
  • 114. T Two β€˜crossed’ polaroids 𝐴 and 𝐡 are placed in the path of a light beam. In between these, a third polaroid 𝐢 is placed whose polarization axis makes an angle πœƒ with the polarization axis of the polaroid 𝐴. If the intensity of light emerging from the polaroid 𝐴 is πΌπ‘œ, then the intensity of light emerging from polaroid 𝐡 will be Solution: Intensity of light emerging from polaroid 𝐢: Intensity of light emerging from polaroid 𝐡: 𝐼𝐡 = 𝐼𝐢 cos2 90Β° βˆ’ πœƒ 𝐼𝐡 = 𝐼0 cos2 πœƒ . cos2 90Β° βˆ’ πœƒ 𝐼𝐡 = 𝐼0 cos2 πœƒ . sin2 πœƒ {Applying the law of Malus.} = 𝐼0 4 2 sin πœƒ cos πœƒ 2 𝐼𝐢 = 𝐼0 cos2 πœƒ 𝐼𝐡 = 𝐼0 4 sin2 (2πœƒ)
  • 115. Unpolarized light with amplitude 𝐴0 passes through two polarizers. The first one has an angle of 30Β° clockwise to vertical and second one has an angle of 15Β° counter-clockwise to the vertical. What is the amplitude of the light emitted from the second polarizer? Given: To find: 𝐴 = 𝐴0 𝐴2 Solution: We know that: 𝐼 ∝ 𝐴2 . ∴ 𝐴 ∝ 𝐼 β‡’ 𝐴1 𝐴2 = 𝐼1 𝐼2 𝐴1 𝐴0 = 𝐼1 𝐼0 = 𝐼0 2𝐼0 = 1 2 𝐴1 = 𝐴0 2 𝐼1 = 𝐼0 2 T 𝐴1 = 𝐴0 2 𝐼2 = 𝐼1cos2 πœƒ 𝐼2 = 𝐼0 2 cos2 45Β° = 𝐼0 2 1 2 2 = 𝐼0 4 𝐴2 𝐴0 = 𝐼2 𝐼0 = 𝐼0 4𝐼0 = 1 4 𝐴2 = 𝐴0 2 𝐴1 𝐴2 = 𝐼1 𝐼2 Again, πœƒ = 30Β° βˆ’ βˆ’15Β° = 45Β°
  • 116. Medium(πœ‡) 𝑖 Oscillations βŠ₯ to the incidence plane Oscillations in the incidence plane Polarization by Reflection ● The reflected light has more vibrations perpendicular to plane of incidence. ● The refracted light has more vibration parallel to the plane of incidence. ● The percentage of polarization in reflected light changes as we change the angle of incidence 𝑖.
  • 117. ● For a particular angle of incidence 𝑖𝐡 , the reflected light becomes completely plane polarized. ● The required condition for this purpose is: 𝑖𝐡 + π‘Ÿ = 90Β° ● Brewster’s Law tan 𝑖𝐡 = πœ‡ 𝑖𝐡 = Brewster angle/polarising angle Brewster’s Law tan 𝑖𝐡 = πœ‡2 πœ‡1 ● If the light ray travels from one medium to another with refractive πœ‡1 and πœ‡2 respectively, then Brewster’s law becomes, Medium(πœ‡) 𝑖𝐡 π‘Ÿ
  • 118. Relation between Critical Angle and Brewster Angle ∴ tan 𝑖𝐡 = 1 sin πœƒπΆ ● Brewster’s Law tan 𝑖𝐡 = πœ‡ 𝑖𝐡 = Brewster angle/polarising angle sin πœƒπΆ = 1 πœ‡ ● Critical Angle: 𝑖𝐡 = tanβˆ’1 1 sin πœƒπΆ And πœƒπΆ = sinβˆ’1 1 tan 𝑖𝐡 Medium(πœ‡) 𝑖𝐡 π‘Ÿ
  • 119. T The polarizing angle of diamond is 67Β°. The critical angle of diamond is nearest to: [Given tan 67Β° = 2.36 ] To find: Solution: Critical angle Given: 𝑖𝐡 = 67Β° Critical angle is given by: πœƒπΆ = sinβˆ’1 1 2.36 1 2.36 < 1 2 β‡’ sinβˆ’1 1 2.36 < sinβˆ’1 1 2 β‡’ πœƒπΆ < 30Β° ∴ Out of the four option only 22Β° is less than 30Β° πœƒπΆ = sinβˆ’1 1 tan 𝑖𝐡 πœƒπΆ = sinβˆ’1 1 tan 67Β°
  • 120. Scattering by Polarization ● If an unpolarized light gets scattered from air molecule, light perpendicular to original ray is plane polarized. ● If we draw a plane perpendicular to the incident light, then from every viewpoint on the plane we can see the plane polarized light. Air molecule Incident Light Observer Observer
  • 121. Given: To find: Solution: 𝑖𝑝 = 60Β° Velocity of light in glass (𝑣𝑔) As reflected light is plane polarized, 𝑖𝑝 = 60∘ According to Brewster’s law, πœ‡ = tan 𝑖𝑝 = tan 60∘ = 3 As, πœ‡ = π‘£π‘Ž 𝑣𝑔 β‡’ 𝑣𝑔 = π‘£π‘Ž πœ‡ T 𝑣𝑔 = 3 Γ— 108 3 = 3 Γ— 108 π‘š/𝑠 A ray of light, travelling in air, is incident on a glass slab with angle of incidence 60Β°. It is found that the reflected ray is plane polarized. The velocity of light in the glass is: Air Plane polarized Glass (πœ‡) 90Β° 60Β°