Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Successfully reported this slideshow.

Like this presentation? Why not share!

4,254 views

Published on

No Downloads

Total views

4,254

On SlideShare

0

From Embeds

0

Number of Embeds

2

Shares

0

Downloads

194

Comments

4

Likes

10

No notes for slide

- 1. PART 3 Wave Optics1. Huygen’s Principle2. Young’s Two Slits Experiment3. Air Wedge4. Interference of Thin Film5. Diffraction by a Single Slit6. Diffraction Grating7. Spectrometer and Spectroscopy8. Polarization
- 2. Christian Huygens• 1629 – 1695• Best known for contributions to fields of optics and dynamics
- 3. Huygen’s Principle
- 4. Huygen’s Principle, Cont.• All points on a given wave front are taken as point sources for the production of spherical secondary waves, called wavelets, which propagate in the forward direction with speeds characteristic of waves in that medium.
- 5. Wave Optics• The wave nature of light is needed to explain various phenomena. – Interference – Diffraction – Polarization
- 6. Conditions for Interference• For sustained interference between two sources of light to be observed, there are two conditions which must be met. – The sources must be coherent. • The waves they emit must maintain a constant phase with respect to each other. – The waves must have identical wavelengths.
- 7. Resulting Interference Pattern• The light from the two slits form a visible pattern on a screen.• The pattern consists of a series of bright and dark parallel bands called fringes.• Constructive interference occurs where a bright fringe appears.• Destructive interference results in a dark fringe.
- 8. Fringe Pattern Young’s Double Experiment• The bright areas represent constructive interference.• The dark areas represent destructive interference.
- 9. Interference Patterns• Constructive interference occurs at the center point.• The two waves travel the same distance. – Therefore, they arrive in phase.
- 10. Interference Patterns, 3• The upper wave travels one-half of a wavelength farther than the lower wave.• The trough of the bottom wave overlaps the crest of the upper wave.• This is destructive interference. – A dark fringe occurs.
- 11. Geometry of Young’s Double Slit Experiment x
- 12. Interference Equations, 2• For a bright fringe• d sin θbright = m λ – m = 0, 1, 2, … – m is called the order number. • When m = 0, it is the zeroth order maximum. • When m = 1, it is called the first order maximum.
- 13. Interference Equations, 3• When destructive interference occurs, a dark fringe is observed.• This needs a path difference of an odd half wavelength.• d sin θdark = (m + ½) λ m = 0, 1, 2, …
- 14. Interference Equations, Final• For bright fringes λL x = m , m = 0,1,2... d bright• For dark fringes λL 1 x = ( m + ), m = 0,1,2... d 2 dark
- 15. Interference Equations, Final• For fringe separation λL y = d separation
- 16. Interference Equations, Final
- 17. Example 1A screen is separated from a double-slitsource by 1.2m. The distance between thetwo slits is 0.030 mm. The second orderbright fringe is measure to be 4.50 cm fromthe centerline. Determine:a) the wavelength of the lightb) the fringe separation
- 18. Example 2In a young’s double slits experiment, aninterference pattern is formed on a screen1.0 m away from the double slits. The doubleslits are separated by 0.25 mm and thewavelength of light used is 550 nm.Determine the distance from the centralbright fringe the distance ofa) the 5th bright fringeb) the 3rd dark fringe
- 19. Air Wedge
- 20. Air Wedge
- 21. Air WedgeSeparation between bright fringes 1 L λ L ≅ ny y= 2H n is number of bright fringeAngular size of air wedge H tan θ = L 1λ = 2y
- 22. Example 3Air wedge is formed by placing a piece of thin paperat the edges of a pair of glass plates. Light ofwavelength 600 nm is incident normally onto theplates. Fringes are observed with fringe separation of0.25 mm. The length of the air wedge is 5.0 cm.a)Determine of the thickness of the piece of paper.b)Estimate the number of bright fringes that areformed on the plate.
- 23. Example 4Figure below shows how a piece of material ofthickness 5μm forms an air wedge between a pair ofa glass plates A and B. Light of wavelength 500 nmis incident normally onto the plates. Estimate thenumber of bright fringes produced.
- 24. Interference in Thin Films• Interference effects are commonly observed in thin films. – Examples are soap bubbles and oil on water• Varied colours observed when incoherent light is incident on the water.• The interference is due to the interaction of the waves reflected from both surfaces of the film.
- 25. Interference in Thin Films1 Phase Reversal 0 or 2 Phase Reversal
- 26. Problem Solving with Thin Films Equation 1 phase 0 or 2 phase m = 0, 1, 2, … reversal reversals2nt = (m + ½) λ constructive destructive2nt = m λ destructive constructive
- 27. Interference in Thin Films• Identify interference in the film(constructive/destructive)• Starts the number of order (m) with m=0 (otherwise)• Find the refraction index of thin film.• Determine the number of phase reversals. (0,1 or 2)• Using table, use correct formula and column
- 28. Example 5a) Calculate the minimum thickness of a soap- bubble film (n=1.33) that will result in constructive interference in the reflected light if the film is illuminated by light with wavelength 602 nm.b) In (a), if the soap-bubble is on top of a glass slide with n=1.50, find the minimum thickness (nonzero thickness) for constructive interference?
- 29. Example 6Nearly normal incident light of wavelength600 nm falls onto a thin uniform transparentfilm in air. The refractive index of the film is1.34. A very bright ray reflected by the surfaceof the film is observed. Determine theminimum thickness of the film.
- 30. Example 7 Refer the figure, find the minimum thickness of the film that will produce at least reflection at a wavelength of 552 nm.Hint: minimum thickness is refer to destructive interference
- 31. Diffraction• This spreading out of light from its initial line of travel is called diffraction. – In general, diffraction occurs when waves pass through small openings, around obstacles or by sharp edges.
- 32. Fraunhofer Diffraction• Fraunhofer Diffraction occurs when the rays leave the diffracting object in parallel directions. – Screen very far from the slit – Converging lens (shown)• A bright fringe is seen along the axis (θ = 0) with alternating bright and dark fringes on each side.
- 33. Single Slit Diffraction, 3• In general, destructive interference occurs for a single slit of width a is given by: a sin θdark = mλ m = ±1, ±2, ±3, …
- 34. Example 8The angular separation between the two firstorder minima diffraction pattern produced bysingle slit is 200. The incident light has wavelength600 nm. Determine the width of the slit.
- 35. Example 9Parallel light of wavelength 550 nm is incidentnormally upon a single slit, thus forming adiffraction pattern on a screen. The first orderminimum is at angular position of 10. Determinethe width of the slit.
- 36. Diffraction Grating, Cont.• The condition for maxima is – d sin θbright = m λ • m = 0, ±1, ±2, …• The integer m is the order number of the diffraction pattern.• Number per lines per cm = 1/d
- 37. Diffraction Grating, Final• All the wavelengths are focused at m = 0 – This is called the zeroth order maximum• The first order maximum corresponds to m = 1
- 38. Example 10A parallel light beam of wavelength 600 nm isincident normally onto a diffraction. The anglebetween the two first order maxima is 300.Determine the number of lines per cm on thediffraction grating.
- 39. Example 11A parallel light beam of wavelength 540 nm isincident normally onto a diffraction grating whichhas 5000 lines per cm. Determine the maximumnumber of orders of maxima which can be formedon both sides of the central maximum.
- 40. Polarization of Light Waves• This is an unpolarized wave.
- 41. Polarization
- 42. Selective Absorption• The intensity of the polarized beam transmitted through the second polarizing sheet (the analyzer) varies as – I = Io cos2 θ • Io is the intensity of the polarized wave incident on the analyzer. • This is known as Malus’ Law and applies to any two polarizing materials whose transmission axes are at an angle of θ to each other.
- 43. Polarization by Reflection• Brewster’s Law relates the polarizing angle to the index of refraction for the material.• θp may also be called Brewster’s Angle.
- 44. Example 12A plane-polarised light beam has intensity I0. It isincident normally onto a polaroid. If the anglebetween the transmission axis of the polaroid andthe plane of polarisation is 600, determine theintensity of the light beam after it has passedthrough the polaroid. (0.25I0)
- 45. Example 13The index of refraction of a glass plate is 1.52.What is the Brewster’s angle when the plate isin air.

No public clipboards found for this slide

Login to see the comments