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Unformatted text preview: Physics 21 Fall, 2011 Solution to HW-27 33-42 A light ray in air strikes the right-angle prism shown in the figure ( 6 B = 28 ). This ray consists of two different wavelengths. When it emerges at face AB , it has been split into two different rays that diverge from each other by 8 . 50 . (a) Find the index of refraction of the prism for each of the two wavelengths. (a) In order to use Snells Law to find the index of refraction, it is necessary to determine the angle each of these light rays make with a line normal to the surface of the prism. This means working out a little geometry. The angle between the incident ray and the normal to the hypotenuse of the prism is = 90 - 28 = 62 Since the 12 angle is given, we can find the angle between the upper refracted ray and the normal: 1 = 90 - 28 + 12 = 74 . The angle between the normal and the other refracted ray is 2 = 1 + 8 . 5 = 82 . 5 Now we can write Snells Law for these situations as n red sin62 = n air sin74 and n blue sin62 = n air sin82 . 5 , where n red ( n blue ) is the index for the red (blue) light. We arent told the exact colors, but we expect the light that is bent less (more) will be more to the red (blue) end of the spectrum. Now, letting n air = 1 be the index in air, we solve for each index: n red = n air sin( 1 ) sin(62 ) = sin(74 ) sin(62 ) = 1 . 09 n blue = n air sin( 2 ) sin(62 ) = sin(82 . 5 ) sin(62 ) = 1 . 12 33-45 Old photographic plates were made of glass with a light-sensitive emulsion on the front surface. This emulsion was somewhat transparent. When a bright point source is focused on the front of the plate, the developed photograph will show a halo around the image of the spot. If the glass plate has a thickness, t = 3.60 mm, and the halos have an inner radius, R = 5.02 mm, what is the index of refraction of the glass? (Hint: Light from the spot on the front surface is scattered in all directions by the emulsion. Some of it is then totally reected at the back surface of the plate and...
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