Explanation Given info: The refractive index of the material is n . Τhe radius of the curvature of the material is R . The light parallel to the base of the material strikes it from a height L . The emergent ray makes an angle θ with the material. Let the angle of incidence is γ as shown in figure below. Figure (I) From the Figure (1) from triangle O P Q , sin γ = L R γ = sin − 1 ( L R ) From the Figure (1) from triangle O P Q , cos γ = ( R 2 − L 2 ) R From Snell’s Law n 1 sin γ = n 2 sin ϕ (1) Here, n 1 is the refractive index of the vacuum. n 2 is the refractive index of material. γ is the angle of incidence. ϕ is the angle of refraction at point P. Substitute 1.00 for n 1 , n for n 2 in equation (1). ( 1.00 ) sin γ = n sin ϕ sin ϕ = sin γ n Substitute L R for sin γ in the above equation. sin ϕ = L n R ϕ = sin − 1 ( L n R ) The value of cos ϕ is , cos ϕ = 1 − sin 2 ϕ Substitute L n R for sin ϕ in the above equation. cos ϕ = n 2 R 2 − L 2 n R From figure (1) in triangle OPS , ϕ + ( α + 90.0 ° ) + ( 90.0 ° − γ ) = 180 α = γ − ϕ Apply Snell’s law at the point S. n 2 sin θ = n 1 sin α Here, n 2 is the refractive index of vacuum. n 1 is the refractive index of the material. Substitute 1.00 for n 2 , n for n 1 , γ − ϕ for α in the above equation. 1.00 sin θ = n sin ( γ − ϕ ) sin θ = n sin ( γ − ϕ ) (2) From trigonometry, the formula of sin ( γ − ϕ ) is, sin ( γ − ϕ ) = sin γ cos ϕ − cos γ sin ϕ Substitute sin γ cos ϕ − cos γ sin ϕ for sin ( γ − ϕ ) in equation (2). sin θ = n ( sin γ cos ϕ − cos γ sin ϕ ) Substitute L R for sin γ , ( R 2 − L 2 ) R for cos γ , L n R for sin ϕ n 2 R 2 − L 2 n R for cos ϕ in the above equation

Physics for Scientists and Engineers with Modern Physics

10th Edition

ISBN: 9781337553292

Author: Raymond A. Serway, John W. Jewett

Publisher: Cengage Learning

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Snell’s law is a formula that gives a relation between the angle of incidence and angle of refraction and is given as:

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Chapter 34, Problem 43AP

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The relation between the emergent angle and n , R and L .

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Chapter 34, Problem 42AP

Chapter 34, Problem 44AP

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Physics for Scientists and Engineers with Modern Physics

Ch. 34.3 - Prob. 34.1QQ Ch. 34.4 - If beam is the incoming beam in Figure 34.10b,...Ch. 34.4 - Light passes from a material with index of...Ch. 34.6 - Prob. 34.4QQ Ch. 34.7 - Prob. 34.5QQ Ch. 34 - In an experiment to measure the speed of light...Ch. 34 - Prob. 2P Ch. 34 - As a result of his observations, Ole Roemer...Ch. 34 - Prob. 4P Ch. 34 - You are working for an optical research company...

Ch. 34 - Prob. 6P Ch. 34 - Prob. 7P Ch. 34 - Prob. 8P Ch. 34 - Prob. 9P Ch. 34 - A ray of light strikes a flat block of glass (n =...Ch. 34 - Prob. 11P Ch. 34 - Prob. 12P Ch. 34 - Prob. 13P Ch. 34 - Prob. 14P Ch. 34 - When you look through a window, by what time...Ch. 34 - Prob. 16P Ch. 34 - You have just installed a new bathroom in your...Ch. 34 - Prob. 18P Ch. 34 - Prob. 19P Ch. 34 - Prob. 20P Ch. 34 - Prob. 21P Ch. 34 - A submarine is 300 m horizontally from the shore...Ch. 34 - Prob. 23P Ch. 34 - Prob. 24P Ch. 34 - Prob. 25P Ch. 34 - Prob. 26P Ch. 34 - Prob. 27P Ch. 34 - Prob. 28P Ch. 34 - Prob. 29P Ch. 34 - Prob. 30P Ch. 34 - An optical fiber has an index of refraction n and...Ch. 34 - Prob. 32AP Ch. 34 - How many times will the incident beam in Figure...Ch. 34 - Prob. 34AP Ch. 34 - Prob. 35AP Ch. 34 - Prob. 36AP Ch. 34 - Prob. 37AP Ch. 34 - Prob. 38AP Ch. 34 - Prob. 39AP Ch. 34 - A light ray enters the atmosphere of a planet and...Ch. 34 - Prob. 41AP Ch. 34 - Prob. 42AP Ch. 34 - Prob. 43AP Ch. 34 - Prob. 44AP Ch. 34 - Prob. 45AP Ch. 34 - Prob. 46AP Ch. 34 - Prob. 47AP Ch. 34 - Prob. 48AP Ch. 34 - Prob. 49AP Ch. 34 - Figure P34.50 shows a top view of a square...Ch. 34 - Prob. 51AP Ch. 34 - Prob. 52CP Ch. 34 - Prob. 53CP Ch. 34 - Pierre de Fermat (16011665) showed that whenever...Ch. 34 - Prob. 55CP Ch. 34 - Prob. 56CP Ch. 34 - Prob. 57CP

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