Question
Asked Dec 25, 2019
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A transparent cylinder of radius R = 2.00 m has a mirrored
surface on its right half, as shown in Figure P22.55. A light ray

traveling in air is incident on the left side of the cylinder. The
incident light ray and the exiting light ray are parallel, and
d = 2.00 m. Determine the index of refraction of the material.

Incident ray
R.
Mirrored
surface
Exiting ray
Figure P22.55
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Incident ray R. Mirrored surface Exiting ray Figure P22.55

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Expert Answer

Step 1

Given:

radius = 2 m

distance between rays = 2m

angle of incidence = θ

angle of refraction = α

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Incident ray d/ 2 >B Mirrored d/ 2 surface Exiting ray - п Figure P22.55

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Step 2

from triangl...

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d/2 sin e = 2/2 sin e 2 sin o 0 = 30° from figure B+0 =180° B+30° = 180° B =150° AC=BC= radius of the circle therefore, ABC is anisosceles triangle from isosceles triangle ABC a+B+y = 180° a+150+a = 180° 2a+150 = 180° a = 15° refractive index of the material sin e sin a sin 30° sin15° n=1.932

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