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# Optics tutorial 1st year physics classes 2013-2014 { Problems n Solutions}

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About 39 questions and their solutions in optics dedicated to 1st class physics students

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### Optics tutorial 1st year physics classes 2013-2014 { Problems n Solutions}

1. 1. 1. A plane wave of wavelength is incident on a barrier in which there is an opening of diameter d. (a) When , the rays continue in a straight-line path, and the ray approximation remains valid. (b) When , the rays spread out after passing through the opening. (c) When , the opening behaves as a point source emitting spherical waves. 2. For light and electromagnetic waves to reach the earth from the distant galaxies, it takes a long period of time to do so even the fact that such electromagnetic waves travel at the fastest speed ever known ( ). Most galaxies are located at distances measured in “Light year” from the earth. Light-Year, in astronomy, unit of length sometimes used to measure vast distances. It is equivalent to the distance that light travels in a mean solar year. At the rate of approximately 300,000 km/sec (186,000 mi/sec), a light-year is equal to 9,461,000,000,000 km (5,880,000,000,000 mi). For example, light reaching us from the sun (at 149,600,000 km away from earth) takes about 8 minutes to traverse the sun – earth distance. In so doing, this light is not of the current moment but represents the past eight minutes of the sun. 3. a) The frequency f of the He-Ne red laser of wavelength is calculated as follows: b) When light travels in air or vacuum it has a wavelength and speed c, but once it inters a medium of different optical density such as glass which is more dense than air, its speed is slowed down and becomes v with wavelength . The frequency remains fixed. The ratio n=c/v is the refractive index of the medium. Therefore, the wavelength is given by: c) The speed v is calculated as follows: ⁄ ⁄ 4. The refractive index of the crown glass is 1.52 and to calculate the angle of reflection one needs to find the angle of incidence . By applying the Snell’s law of refraction at the air – crown glass media interface ever since the angle of refraction Is given, the angle of incidence can be evaluated: Baghdad University – College of Science – Physics Department st Optics Tutorial – 1 Classes December 25, 2013 1 Dr. Qahtan Al-zaidi Mobile: +9647702981421
2. 2. According to the law of reflection, the reflected ray, the reflected ray, and the normal all lie in the same plane 5. a) The refractive index of the corn syrup n2 is calculated by applying Snell law of refraction at the water – corn ( ) syrup interface as follows: b) The wavelength in the medium is lesser than that in the vacuum: c) The frequency of light as it enters the medium remains unchanged (fixed always for the particular light) and is calculated as: d) The speed of light v in the medium is: 6. a) The speed of light ( )( ) in the flint glass medium is: b) In water, n=1.33 so light speed will be: c) For cubic zirconia, the refractive index is 2.2 and the speed of light waves will be: Baghdad University – College of Science – Physics Department st Optics Tutorial – 1 Classes December 25, 2013 2 Dr. Qahtan Al-zaidi Mobile: +9647702981421
3. 3. 7. Applying Snell’s law of refraction at the water – transparent substance interface: [ ][ ] 8. Applying Snell’s law of refraction at the first air – glass interface to find the angle of refraction as follows: [ From the figure at the right, applying Snell’s law at the second glass – air interface: ] [ ] 𝜃𝑖 𝑛 𝑎𝑖𝑟 𝑛 𝑔𝑙𝑎𝑠𝑠 𝜃𝑡 [ ] [ ] It is obvious that the light enters and exits the flat block of glass with the same angle of 30 degrees with the normal to the surface. However, the direction of propagation of light rays is displaced by a distance d with respect to the incident rays. 𝜃𝑖 𝜃𝑡 d 9. A transverse wave is one in which the elements of the medium move in a direction perpendicular to the direction of propagation. An example is a wave on a taut string. Also, the electromagnetic waves are transverse. Longitudinal wave is one in which the elements of the medium move in a direction parallel to the direction of propagation. Sound waves in fluids are longitudinal. 10. Sound waves Sound waves are longitudinal in which the elements of the medium move in a direction parallel to the direction of propagation. Sound waves and travel through a compressible medium with a speed that depends on the elastic and inertial properties of that medium. By definition, the propagation of mechanical disturbances—such as sound waves, water waves, and waves on a string—requires the presence of a medium. Light waves Light waves are transverse, i.e., the components of the electric and magnetic fields of plane electromagnetic waves are perpendicular to each other and perpendicular to the direction of wave propagation. Electromagnetic waves (unlike mechanical waves) can travel through empty space at the speed of light c The waves carry energy. The rate of flow of energy crossing a unit area is described by the Poynting vector S, Electromagnetic waves carry momentum and hence exert pressure on surfaces. The electromagnetic spectrum includes waves covering a broad range of wavelengths, from long radio waves at more than 104 m to gamma rays at less than 10-14 m. 11. Electromagnetic waves are generated by oscillating electric charges. The waves consist of oscillating electric and magnetic fields at right angles to each other and to the direction of wave propagation. Thus, electromagnetic waves are transverse waves. The waves radiated from the oscillating charges can be detected at great distances. Furthermore, electromagnetic waves carry energy and momentum and hence can exert pressure on a surface. Baghdad University – College of Science – Physics Department st Optics Tutorial – 1 Classes December 25, 2013 3 Dr. Qahtan Al-zaidi Mobile: +9647702981421
4. 4. 12. Astonishingly, Maxwell’s equations also predict the existence of electromagnetic waves that propagate through space at the speed of light c. The Ampère–Maxwell law predicts that a time-varying electric field produces a magnetic field, just as Faraday’s law tells us that a time-varying magnetic field produces an electric field. Therefore, not all current – carrying conductors are able to emit electromagnetic waves. Only when there is a time – varying of current in a conductor there is an electromagnetic wave. 13. Radio waves (frequencies of about 107 Hz) are electromagnetic waves produced by oscillating currents in a radio tower’s transmitting antenna. Light waves are a high-frequency form of electromagnetic radiation (about 1014 Hz) produced by oscillating electrons in atoms. 14. The magnitudes of E and B in empty space are related by the expression: Electric and magnetic fields are measured in different units, however, so they cannot be directly compared. The electric and magnetic fields contribute equally to the energy of the wave. That is, the instantaneous energy density associated with the magnetic field of an electromagnetic wave equals the instantaneous energy density associated with the electric field. Hence, in a given volume the energy is equally shared by the two fields. 15. The frequency f (in Hz = cycles/second) of the electromagnetic wave can be calculated as given below: 16. 17. a) For a person of 175 - cm average tall: b) For a typical paper of 0.1 mm, the wavelength and frequency of such waves will be: Baghdad University – College of Science – Physics Department st Optics Tutorial – 1 Classes December 25, 2013 4 Dr. Qahtan Al-zaidi Mobile: +9647702981421
5. 5. 18. a) The wavelength follows: of electromagnetic wave in free space of frequency b) For electromagnetic wave of frequency is evaluated as , the wavelength is calculated as: 19. Electromagnetic waves, including light, travel in free space with speed of where is the wavelength in vacuum and is the frequency. When propagating in medium other than free space, light travels at a lower speed of As light travels from one medium to another, its frequency does not change but its wavelength does. To see why this is so, consider the Figure to the right. Waves pass an observer at point A in medium 1 with a certain frequency and are incident on the boundary between medium 1 and medium 2. The frequency with which the waves pass an observer at point B in medium 2 must equal the frequency at which they pass point A. If this were not the case, then energy would be piling up at the boundary. Because there is no mechanism for this to occur, the frequency must be a constant as a light ray passes from one medium into another. Therefore, because the relationship must be valid in both media and because , we see that Because , it follows that . 20. a) The frequency So, the number of wavelengths you are away from the station will be b) For the FM band, the frequency is and this is equivalent to a wavelength of 21. For people sitting next to their radios 100 km away from the station, the radio waves take time to reach them, while the people sitting across the room will hear the sound waves from the newscaster after a time . Apparently, the far away people receive the news first. 22. The shadow length on the vertical wall is: ( [ ] ) ( ) The same shadow length results when the halving the distance between the point source and the edges of the meter stick since this causes to half the angle and consequently, tan . Baghdad University – College of Science – Physics Department st Optics Tutorial – 1 Classes December 25, 2013 5 Dr. Qahtan Al-zaidi Mobile: +9647702981421
6. 6. 100 cm 𝜃 582 cm 264 cm 23. The frequency f of the yellow sodium light is: The wavelength in air (or vacuum) changes to when propagating in the medium. For water, it becomes: 24. When light enters a medium, its speed reduces to glass sheet is calculated by: . Here, ( ⁄ ) 25. The energy of an orange light photon of wavelength ( )( ( =(2/3) c and hence the refractive index n of the (Not 670 nm !!) is calculated as follows: ) ) 26. 27. 28. 29. 30. The shapes are the same, but (a) is a graph of vertical position versus horizontal position while (b) is vertical position versus time. Figure (a) is a pictorial representation of the wave for a series of particles of the medium— this is what you would see at an instant of time. Figure (b) is a graphical representation of the position of one element of the medium as a function of time. The fact that both figures have the identical shape represents Equation ( ) ( ) a wave is the same function of both x and t. 32. See solution of question 35. 33. The critical angle is the angle of incidence of light in the more dense medium (higher refractive index n1) at which light refracts at an angle of 90 o to the normal in the lesser dense medium (lower refractive index n2). Thus, the critical angle a) For diamond, n=2.419, b) For flint glass, n=1.660, c) For ice of n=1.309, 34. If the rarer medium is water (n=1.33) other than air (n=1), then the critical angle replacing the previous value of n2=1 by its value for water, i.e., n2=1.33 as follows: will be calculated by a) For diamond, n=2.419, b) For flint glass, n=1.660, c) For ice of n=1.309 surrounded by water of n=1.33, no critical angle can be found!! Why?? Baghdad University – College of Science – Physics Department st Optics Tutorial – 1 Classes December 25, 2013 6 Dr. Qahtan Al-zaidi Mobile: +9647702981421
7. 7. 35. The maximum refraction angle at which light ray can emerge from the other side of the prism is 90 degrees to the normal which occurs when light incident at the critical angle given by: a b ( ( ( 60 o f h ) )) θ d [ i b ] g c b 42 o e d c 90 o h g This is the smallest incident angle above which light will not emerge from the other side of the prism and light will totally internally reflected TIR. 36. The distance d that the direction of propagation of light ray will be displaced is given by: [ ( ] [ ] ( ) ) 37. The time interval required for the light to pass through the glass block described in the previous problem is simply given by: ( )( ) 38. For a typical glass window of average width of 6 mm, the delay of time results from light being pass through this glass window is calculated as: ( )( ) ( 39. The difference between the two mean deviation angles, such as the spectrum. That is: ( ) ( ) ( ) To calculate the dispersive power flint glass. Thus: is a measure of the dispersion of ( )( ) one needs to evaluate the index of refraction for yellow light [ Therefore, the dispersive power ) ] [ in silica ] is evaluated as follows: ( ( ) ) ( ( ) ) Baghdad University – College of Science – Physics Department st Optics Tutorial – 1 Classes December 25, 2013 7 ( ( ) ) Dr. Qahtan Al-zaidi Mobile: +9647702981421