Question

Chapter 33, Problem 051 In the figure, light is incident at angle 01-37.0° on a boundary between two transparent materials. Some of the light travels down through the next three layers of transparent materials, while some of it reflects upward and then escapes into the air. If n1-1.28, n2-1.36, n): 1.34 and n4-1.47, what is the value of (a) es and (b) θ? Air 21 ns 24 (a) 6sNumber (b) 64 Click if you would like to Show Work for this question: Units Number Units Open Show Work

Answer 1

Apply Snell's Law for the interface of first two transparent materials,

n₁sin $\theta$ ₁ = n₂sin $\theta$ ₂

1.28 * sin37 = n₂ * sin $\theta$ ₂

0.7703 = n₂*sin $\theta$ _{2 .........}(1)

Apply Snell's Law for the interface of second and third material,

n₂ sin $\theta$ ₂ = n₃ sin $\theta$ ₃

From eq (1),

0.7703 = n₃ * sin $\theta$ ₃ ........(2)

Apply Snell's Law for the interface of third and fourth material,

From eq (2),

n₃sin $\theta$ ₃ = n₄sin $\theta$ ₄

0.7703 = 1.47 * sinθ₄

sin $\theta$ ₄ = 0.524

$\theta$ ₄ = 31.60 deg

Apply Snell's Law for the interface of first two transparent material and air,

n₁ sin $\theta$ ₁ = n_a sin $\theta$ ₅

0.7703 = 1 * sin $\theta$ ₅

$\theta$ 5 = 50.38 deg