Question
Asked Dec 25, 2019
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Determine the maximum angle u
for which the light rays incident on
the end of the light pipe in Figure
P22.38 are subject to total internal
reflection along the walls of the
pipe. Assume the light pipe has an
index of refraction of 1.36 and the
outside medium is air.

|2.00 μη
γ
νL
Figure P22.38
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|2.00 μη γ νL Figure P22.38

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Given that the refractive index of air is 1.00, and the refractive index of light pipe is 1.36. The expression for critical angle is 0. п, 0. = sin air Пруре gipe 1.00 0. = sin 1.36 [1] 0. = 47.3° The angle of refraction is given by 0. = 90° – 0. Use equation [1]in the above equation to find 0,. 0, = 90° – 47.3° [2] 0, = 42.7° %3D

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