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

To determine

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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Chapter 34 Solutions

Physics for Scientists and Engineers with Modern Physics

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