Question

A block of crown glass of refractive index 1.52, is immersed in water of refractive index 1.33, as in the figure shown below. A light ray is incident on the top face at an angle of θ₁ = 42.2° with the normal and exits the block at point P. (Assume that x = 3.63 cm.)

A block is immersed in water. A ray starts in the water at the top of the image and moves down and to the right before it is incident on the top surface of the block at an angle θ₁ to the vertical at a distance xfrom the right side of the block. The ray refracts to have a steeper slope than the incident ray, and extends down and to the right to point P on the right side of the block, where point P is a vertical distance y from the top of the block. The ray exits the block at a steeper angle than it was within the block, and at an angle θ₂ to the horizontal.

(a) Find the vertical distance y from the top of the block to P.
cm

(b) Find the angle of refraction θ₂ of the light ray leaving the block at P.
°

I WILL RATE

Answer 1

a)

applying Snell’s law of refraction at the top surface,

n1 Sinθ1 = n2 Sinθ

1.33 x Sin 42.2 = 1.52 x Sinθ

Angle of refraction at the top surface, θ = 36⁰

Tan θ = x/y

Vertical distance, y = x/tan θ = 3.63/tan 36 = 5.0 cm

The vertical distance, y = 5.0 cm

b)

the angle of incidence on the vertical side at P is I = 90 – θ = 90 – 36 = 54⁰

applying Snell’s law of refraction at the vertical surface at P,

n2 Sin i = n1 Sinθ2

1.52 x Sin 54 = 1.33 x Sinθ2

Angle of refraction at the vertical surface, θ2 = 67.6⁰