Chapter 17: Wave Optics Solutions

Transcription

1 Chapter 17: Wave Optics Solutions Questions: 3, 7, 11, 15 Exercises & Probles: 4, 7, 8, 10, 1, 4, 44, 55 Q17.3: The wavelength of a light wave is 700 n in air; this light appears red. If this wave enters a pool of water, its wavelength becoes λ air /n = 530 n. If you were swiing underwater, the light would still appear red. Given this, what property of a wave deterines its color? Explain. ( λ Q17.3. Reason: The wavelength changes n λair/ n) when the light goes fro one ediu to another, but the frequency stays the sae. Since the color doesn t change, then it ust be associated with the frequency. Assess: Energy of the light is directly related to the frequency as well. = Q17.7: The figure shows the viewing screen in a double-slit experient with onochroatic light. Fringe C is the central axiu. a. What will happen to the fringe spacing if the wavelength of the light is decreased? b. What will happen to the fringe spacing if the spacing between the slits is decreased? c. What will happen to the fringe spacing if the distance to the screen is decreased? d. Suppose the wavelength of the light is 500 n. How uch farther is it fro the dot on the screen in the center of fringe E to the left slit than it is fro the dot to the right slit? Q17.7. Reason: The fringe separation for the light intensity pattern of a double slit is y=d /. deterined by We can answer this question by inspecting this relationship. (a) If λ y will decrease. (b) If d y will increase. (c) If L y will decrease. (d) Since the dot is in the = bright fringe, the path length difference fro the two slits is λ = 1000 n. Assess: The ability to inspect a relationship and answer what if questions is a skill that physics will help you develop. Q17.11: Why does light reflected fro peacock feathers change color when you see the feathers at a different angle? Explain. Q Reason: Peacock feathers consist of thin nearly parallel rods of elanin. Since they are thin, thin-fil interference is a factor. Since they are very sall and nearly parallel, they also act as a diffraction grating. As a result different wavelengths add up constructively at different locations and you will see different colors when viewing the feather fro different angles. 1

2 Assess: This effect is coon fro biological structures whose size is siilar to the wavelength of light. Q17.15: Should the antireflection coating of a icroscope objective lens designed for use with ultraviolet light be thinner, thicker, or the sae thickness as the coating on a lens designed for visible light? Explain. Q Reason: The wavelength of ultraviolet light is shorter than the wavelength of visible light. The forulas for the thickness of thin fils, such as those used as antireflection coatings, show the thickness is proportional to the wavelength. Hence, if the wavelength is shorter, then the fil needs to be thinner. Assess: It is difficult to ake antireflection coatings that work at a wide range of wavelengths, but it can be done soewhat with ultiple layers of fils of different indices of refraction. P17.4: A light wave has a 670 n wavelength in air. Its wavelength in a transparent is 40 n. a) What is the speed of light in this? b) What is the light s frequency in the? P17.4. Prepare: Light rays travel in straight lines and light s speed in a aterial is characterized by its refractive index, defined by Equation Also, the frequency does not change as the wave oves fro one ediu to another. Solve: (a) The refractive index is c f λ λ λ 40 n n= = = v = c = = v 670 n vac vac fλ λ λvac or /s to two significant figures. (b) The frequency is f v λ /s 14 = = = Hz 40 n 8 8 ( /s) /s Assess: We ust not forget that the frequency of a wave does not change as the wave oves fro one ediu into another. P17.7: Two narrow slits 50 μ apart are illuinated with light of wavelength 500 n. What is the angle of the = bright fringe in radians? In degrees? P17.7. Prepare: Two closely spaced slits produce a double-slit interference pattern given by Equation The interference pattern looks like the photograph of Figure It is syetrical with the fringes on both sides of and equally distant fro the central axiu. Solve: The bright fringes occur at angles θ such that dsinθ = λ = 0, 1,, 3,... ( ) 180 sin θ = = 0.0 = 0.00 rad = 1.1 (50 10 ) θ = 0.00 rad π rad

3 Assess: We did expect the angle to be sall because the wavelength of light is uch saller than the separation of the two slits. P17.8: Light fro a sodiu lap (λ = 589 n) illuinates two narrow slits. The fringe spacing on a screen 150 c behind the slits is 4.0. What is the spacing (in ) between the two slits? P17.8. Prepare: Two closely spaced slits produce a double-slit interference pattern. The interference pattern looks like the photograph of Figure Solve: The fringe spacing is given by Equation 17.9 as follows: ( )( ) y = d = = 3 d y = 0. P17.10: A double-slit experient is perfored with light of wavelength 600 n. The bright interference fringes are spaced 1.8 apart on the viewing screen. What will the fringe spacing be if the light is changed to a wavelength of 400 n? P Prepare: Two closely spaced slits produce a double-slit interference pattern. The interference pattern looks like the photograph of Figure Solve: The forula for fringe spacing is Equation λ y = L ( ) L = d d L 3000 d = The wavelength is now changed to 400 n, and Ld /, being a part of the experiental setup, stays the sae. Applying the above equation once again, y = = = d ( )(3000) 1. P17.1: Two narrow slits are 0.1 apart. Light of wavelength 550 n illuinates the slits, causing an interference pattern on a screen 1.0 away. Light fro each slit travels to the = 1 axiu on the right side of the central axiu. How uch farther did the light fro the left slit travel than the light fro the right slit?? P17.1. Prepare: For a double slit, constructive interference occurs when r = λ where = 0, 1,, 3,... Solve: At the first interference axiu ( = 1) the path difference fro the two slits is r = λ = (1) λ = 550 n. one wavelength, which in this case is Assess: For a double slit, the first interference axiu occurs when the path difference fro the two slits is one wavelength. P17.4: Antireflection coatings can be used on the inner surfaces of eyeglasses to reduce the reflection of stray light into the eye, thus reducing eyestrain. a) A 90-n-thick coating is applied to the lens. What ust be the coating s index of refraction to be ost effective at 480 n? Assue that the coating s index of refraction is less than that of the lens. 3

4 b) If the index of refracting of the coating is 1.38, what thickness should the coating be so as to be ost effective at 480 n? The thinnest possible coating is best. P17.4. Prepare: First note that the coating is on the inside of the glass. As the incident light reflects off the air-coating interface there is no phase change. As the light that is transitted through the coating reflects off the coating-air interface there is no phase change. Destructive interference for two reflective no-phase changes occurs when t = ( + 1/) λair/ n. We will use = 1 in order to obtain the thinnest possible coating. Solve: (a) The index of refraction for the coating is n= ( + 1/) λair/( t) = λair/(4 t) = 480 n/(4(90 n)) = (b) The thickness of the coating is 1 t = + λair / n= λ/4 n= (480 n)/4(1.38) = 87 n Assess: This is a reasonable value for the index of refraction thickness of the coating (see Proble 17.6). P17.44: The two ost proinent wavelengths in the light eitted by a hydrogen discharge lap are 656 n (red) and 486 n (blue). Light fro a hydrogen lap illuinates a diffraction grating with 500 lines/, and the light is observed on a screen 1.50 behind the grating. What is the distance between the first-order red and blue fringes? P Prepare: A diffraction grating produces an interference pattern. The interference pattern looks like the diagra of Figure The bright interference d sin θ, fringes are given by Equation 17.1 = λ = 0, 1,, 3,... The slit spacing d = = is Solve: θ 1/ and = 1. For the red and blue light, = sin = red θ = sin = y = y y blue 1red 1blue The distance between the fringes, then, is where y y 1red 1blue So, = (1.5 )tan19.15 = 0.51 = (1.5 )tan14.06 = y = = 14.5 c. P17.55: If sunlight shines straight onto a peacock feather, the feather appears bright blue when viewed fro 15 on either side of the incident bea of sunlight. The blue color is due to diffraction fro the fro the elanin bands in the feather barbules. Blue light with a wavelength of 470 n is diffracted at 15 by these bands (this is the first-order diffraction) while other wavelengths in the sunlight are diffracted at different angles. What is the spacing of the elanin bands in the feather? P Prepare: The peacock feather is acting as a reflection grating, so we ay use Equation 17.1: d sin θ, = λ with = 1 λ = 470 n, (we are told it is first-order diffraction), θ = 15 = 0. 6 rad. and 1 Solve: Solve Equation 17.1 for d. 4

5 λ (1)( ) d = = = 18. µ sin θ sin(0. 6 rad) Assess: The answer is sall, but plausible for the bands on the barbules. 5