Question

A quarter cylinder of radius R and refractive index 1.5 is placed on a table. A point object P is kept at a distance mR from it. Find the value of m for which a ray from P will emerge parallel to the table as shown in the figure.

Answer 1

Answer:

Explanation:

=> Refraction at plane surface 1,

n_1 = 1

n_2= 1.5

u = -mR

v = ?

=> Using formula,

n_2/v - n_1/u = n_2-n_1 / R

But, for the plane surface, R = ∞

∵ 3/2 / v - 1/(-mR) = 0

∴ v = -3/2mR

u = -(3/2mR + R) and

R = -R

v = ∞,

n_1 = 3/2,

n_2 = 1

1/∞ - 3/2 / [-(3/2mR + R)] = 1 - 3/2 / -R

=> 3/2 / R(3m + 2) / 2 = 1/2 / R

=> 6 / 3m + 2 = 1

=> 3m + 2 = 6

=> 3m = 6 - 2

=> m = 4/3