Question

light is incident at angle θ1 = 40.1° 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 = 1.30, n2 = 1.40, n3 = 1.32, and n4 = 1.45, what is the value of (a) θ5in the air and (b) θ4 in the bottom material?

Answer 1

Diagram to question is attached.

Answer:

a) 56.87°

b) 35.26°

Explanation:

We are given:

$∅_1 = 40.1$

$n_1 = 1.30$

$n_2 = 1.40$

$n_3 = 1.32$

$n_4 = 1.45$

$n_u = 1$

a) To find ∅$_5$ in air, we use the expression:

$n_1$Sin∅1 = $n_u$Sin∅5

Let's substitute figures in the expression above.

We now have:

1.3 Sin40.1 = 1 Sin∅5

= 0.837 = Sin∅5

= Sin∅5 = 0.837

∅5 = $Sin^-^1 0.837$

∅$_5$ = 56.87°

b) We use the expression:

$n_u$ Sin∅5=n1 Sin1= n2 Sin∅2=n3 Sin∅3=n4 Sin∅4

Since we are to look for ∅4, we have:

n4sin∅4= n_uSin∅5

= 1.45Sin∅4 = 0.837

∅$_4$ = 35.26°

Answer 2

Answer:

a)θ₅ = 56.87°

b)θ₄ = 35.24°

Explanation:

Applying Snell's Law at the interface face of first two transperant materials:

n₁sinθ₁ = n₂sinθ2₂

1.3*sin40.1° = 1.4*sinθ₂

1.4*sinθ₂ = 0.837

sinθ₂ = 0.598

Applying Snell's Law at the interface face of second and third transperant materials:

n₂sinθ₂ = n₃sinθ₃

0.837 = 1.32*sinθ₃

sinθ₃ = 0.634

Applying Snell's Law at the interface face of third and fourth transperant materials:

n₃sinθ₃ = n₄sinθ₄

0.837 = 1.45*sinθ₄

sinθ₄ = 0.577

θ₄ = 35.24°

Applying Snell's Law at the interface face of first transperant material and the air:

n₁sinθ₁ = nairsinθ₅

1.3*sin40.1° = sinθ₅

0.8374 =  sinθ₅

θ₅ = 56.87°