Question

In the drawing, suppose that the angle of incidence is 1 = 42.5°, the thickness of the pane is 3.15 mm, and the refractive index of the pane is n2 = 1.58. Find the amount (in mm) by which the emergent ray is displaced relative to the incident ray.

Answer 1

Given,

n1 = n-air = 1 ; theta1 = 42.5 deg ;

n2 = 1.58 ; d = 3.15 mm

We know from Snell's law

n1 sin(theta1) = n2 sin(theta2)

sin(theta2) = n1 sin(theta1)/n2

sin(theta2) = 1 x sin42.5/1.58 => theta2 = 25.31 deg

Hence, theta2 = 25.31 deg

Again from snell's law

n2 sin(theta2) = n3 sin(theta3)

theta3 = sin^-1(1.58 x sin25.31/1) = 42.5 deg

theta1 = theta3

using trigonometry

x = d sin(theta1 - theta2)

x = 3.15 mm sin(42.5 - 25.31) = 0.931 mm

Hence, x = 0.931 mm