23_InstSolManual_PDF_Part11 - X into the water and then...

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23.42. Set Up: Snell’s law says Let and be the two refractive indicies. Solve: (a) and (b) A ray from part of one letter is shown in Figure 23.42. When it passes into the air it splits into two rays, where each ray has an angle of refraction corresponding to one of the two indicies of refraction. When traced backwards, these two rays appear to come from two different points on the page, corresponding to two images of the letters. Figure 23.42 23.43. Set Up: The light first refracts from air into the glass and then from the glass into the water. The path of the ray is sketched in Figure 23.43. and Snell’s law says Figure 23.43 Solve: For the refraction For the refraction But so and Reflect: The angle of the ray in the water is the same as if the glass plate wasn’t there. 23.44. Set Up: Snell’s law says Apply Snell’s law to the refraction from material
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Unformatted text preview: X into the water and then from the water into the air. Solve: (a) material X to water: and n a 5 n b 1 sin u b sin u a 2 5 1 1.333 2 1 sin 48° sin 25° 2 5 2.34 u b 5 48°. u a 5 25° n b 5 n w 5 1.333. n a 5 n X , n a sin u a 5 n b sin u b . u r b 5 23.4°. sin u r b 5 1 n air n water 2 sin u a 5 1 1.00 1.333 2 sin 1 32.0° 2 5 0.3975 n air sin u a 5 n water sin u r b . u r a 5 u b , n water sin u r b . n glass sin u r a 5 glass S water, n air sin u a 5 n glass sin u b . air S glass, u a u b Air Glass Water u9 a u9 b n a sin u a 5 n b sin u b . u r a 5 u b . n water 5 1.333. n glass 5 1.60 u a 5 32.0°. 1 2 2 1 n 2 5 1 1.00 2 1 sin 63.0° sin 42.1° 2 5 1.33. n 1 5 1 1.00 2 1 sin 63.0° sin 51.3° 2 5 1.14 n b 5 n a 1 sin u a sin u b 2 . n 2 n 1 n a sin u a 5 n b sin u b . Electromagnetic Waves 23-11...
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