23_InstSolManual_PDF_Part22

23_InstSolManual_PDF_Part22 - when is larger. The blue...

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23.82. Set Up: The path of the ray is sketched in Figure 23.82. The ray enters the prism at normal incidence so is not bent. The incident angle at the prism water interface is to be the critical angle. For water, Figure 23.82 Solve: From the figure, gives 23.83. Set Up: The path of the ray is sketched in Figure 23.83. The problem asks us to calculate Figure 23.83 Solve: Apply Snell’s law to the air liquid refraction. and and so Snell’s law applied to the liquid air refraction gives and Reflect: The light emerges from the liquid at the same angle from the normal as it entered the liquid. 23.84. Set Up: The angle of incidence for the glass oil interface must be the critical angle, so Solve: gives 23.85. Set Up: The polarizing angle is given by Solve: (a) Red: Blue: (b) differs more from
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Unformatted text preview: when is larger. The blue light is refracted more upon entering the glass. n b u a u b sin u b 5 n a sin u a n b . n glass 5 tan 70.0 5 2.75. n glass 5 tan 60.0 5 1.73. n glass 5 1 1.00 2 tan u p . n b 5 n glass . n a 5 1.00. tan u p 5 n b n a . u p n oil 5 1 1.52 2 sin 57.2 5 1.28. 1 1.52 2 sin 57.2 5 n oil sin 90. n a sin u a 5 n b sin u b u b 5 90. S u r b 5 42.5. 1 1.00 2 sin u r b 1 1.63 2 sin 1 24.5 2 5 S u r a 5 u b 5 24.5. f 5 u r a , u b 5 f u b 5 24.5. 1 1.00 2 sin 1 42.5 2 5 1 1.63 2 sin u b S u a u b f f u9 a u9 b u r b . n glass 5 1.333 sin 45 5 1.89. n glass sin 45 5 1 1.333 2 sin 90. n a sin u a 5 n b sin u b u crit 5 45. 45 45 u crit n water 5 1.333. S 23-22 Chapter 23...
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