Chapter 35

Chapter 35 - CHAPTER Interference and Diffraction 35 1*...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
CHAPTER 35 Interference and Diffraction 1* When destructive interference occurs, what happens to the energy in the light waves? The energy is distributed nonuniformly in space; in some regions the energy is below average (destructive interference), in others it is higher than average (constructive interference). 2 Which of the following pairs of light sources are coherent: ( a ) two candles; ( b ) one point source and its image in a plane mirror; ( c ) two pinholes uniformly illuminated by the same point source; ( d ) two headlights of a car; ( e ) two images of a point source due to reflection from the front and back surfaces of a soap film. ( b ), ( c ), and ( e ) 3 ( a ) What minimum path difference is needed to introduce a phase shift of 180 ° in light of wavelength 600 nm? ( b ) What phase shift will that path difference introduce in light of wavelength 800 nm? ( a ), ( b ) Use Equ. 35-1 ( a ) 300 nm ( b ) δ = 135 o 4 Light of wavelength 500 nm is incident normally on a film of water 10 –4 cm thick. The index of refraction of water is 1.33. ( a ) What is the wavelength of the light in the water? ( b ) How many wavelengths are contained in the distance 2 t , where t is the thickness of the film? ( c ) What is the phase difference between the wave reflected from the top of the air–water interface and the one reflected from the bottom of the water–air interface after it has traveled this distance? ( a ) λ n = / n ( b ) N = 2 t / n ( c ) = π + 2 N n = 376 nm N = 5.32 = 6.32 rad = 0.32 rad 5* ∙∙ Two coherent microwave sources that produce waves of wavelength 1.5 cm are in the xy plane, one on the y axis at y = 15 cm and the other at x = 3 cm, y = 14 cm. If the sources are in phase, find the difference in phase between the two waves from these sources at the origin. 1. Find r = r 2 r 1 2. Use Equ. 35-1 r 1 = 15 cm, r 2 = 14.318 cm; r = 0.682 cm = (0.682/1.5)360 o = 164 ° 6 The spacing between Newton’s rings decreases rapidly as the diameter of the rings increases. Explain qualitatively why this occurs. The thickness of the air space between the flat glass and the lens is approximately proportional to the square of d , the diameter of the ring. Consequently, the separation between adjacent rings is proportional to 1/ d .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Chapter 35 Interference and Diffraction 7 ∙∙ If the angle of a wedge-shaped air film such as that in Example 35-2 is too large, fringes are not observed. Why? The distance between adjacent fringes is so small that the fringes are not resolved by the eye. 8 ∙∙ Why must a film used to observe interference colors be thin? If the film is thick, the various colors (i.e., different wavelengths) will give constructive and destructive interference at that thickness. Consequently, what one observes is the reflected intensity of white light (see Problem 33-75). 9* A loop of wire is dipped in soapy water and held so that the soap film is vertical. ( a ) Viewed by reflection with white light, the top of the film appears black. Explain why. ( b ) Below the black region are colored bands.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 19

Chapter 35 - CHAPTER Interference and Diffraction 35 1*...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online