Question

Suppose n1 = 1.11, n2 = 1.3 and n3 = 1.35. What is θ3 in degrees if the angle of incidence is 40.8?

Answer 1

given data:

n₁​ = 1.11 , n₂ = 1.3 , n₃​ = 1.35 , angle of incidence , θ_​i ​= θ_​1 = 40.8°

first we understand this case by figure ,

first we find refraction angle ( θ₂​) by using of snell's law,

n₁ sin(θ₁​) = n₂ sin(θ₂​)

1.11 x sin ( 40.8°​) = 1.3 x sin(θ₂​)

1.11 x 0.5979 = 1.3 x sin(θ₂​)

0.6636 / 1.3 = sin(θ₂​)

sin(θ₂​)​ = 0.5105

θ₂​ = sin^-1​( 0.5105 )

θ₂​ =​ 34.10°

here we got value of angle of refraction for medium (1-2) is equal to θ₂​ =​ 34.10°

now , from figure we can says that angle of refraction θ₂​ become angle of incidence for medium (2-3)

for medium (2-3) ,

n₂ sin(θ₂​) = n₃ sin(θ₃)

1.3 x sin(34.10°) = 1.35 x sin(θ₃)

1.3 x 0.510 = 1.35 x sin(θ₃)

0.663 = 1.35 x sin(θ₃)

0.663 / 1.35 = sin(θ₃)

0.4911 = sin(θ₃)

θ₃ = sin^​-1( 0.4911)

θ₃ = 32.68°