Explanation Given Info: The radius of a transparent cylinder is 2.00 m , the distance between the incoming ray and the outgoing ray is 2.00 m . The diagram that represents a transparent cylinder having a mirrored surface in its right half as shown in figure. Figure (1) From the figure above in triangle Δ P S C , sin θ = ( d / 2 ) R Rearrange the above expression for the θ . θ = sin − 1 ( ( d / 2 ) R ) Substitute 2.00 m for d and 2.00 m for R in above equation to find θ . θ = sin − 1 ( ( 2.00 m / 2 ) 2.00 m ) = 30 ° Th path must be symmetric about the centerline C Q of the cylinder if the emerging ray is parallel to the incident ray. The triangle Δ P Q C is an isosceles triangle so the θ 3 and θ 1 are equal because angle opposite to equal sides are equal. θ 3 = θ 1 The angle θ and θ 2 makes an linear pair. So the sum of both of them is 180 ° . Write the expression for the angle θ and θ 2 . θ + θ 2 = 180 ° θ 2 = 180 ° − θ Write the expression for the angle θ 1 , θ 2 and θ 3 for a triangle Δ P Q C in which the sum of all the angles of the triangle is 180 °

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ISBN: 8220100663987

Author: Jewett

Publisher: Cengage Learning US

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Chapter 35, Problem 35.74AP

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The index of refraction of the material.

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Chapter 35, Problem 35.73AP

Chapter 35, Problem 35.75AP

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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 (page 800). 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.

A transparent cylinder of radius R = 2.00 m has a mirroredsurface 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. Theincident light ray and the exiting light ray are parallel, andd = 2.00 m. Determine the index of refraction of the material.

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Chapter 35 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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The index of refraction of the material.

Solution Summary: The author explains that the index of retion of the material is 1.93.

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Chapter 35 Solutions