Answer to Question #97335 in Optics for Oluwole Henry

Question

A thin equivalent lens is placed on a horizontal plane mirror and a pin held 20cm vertically above the lens coincise in position with its own image. The image between the under surface of the lens and the mirror is spilled with water (refractive index 1.33), then to coincise with its image, the pin had to be raised until its distance from the lens is 27.5cm. Find the radius of the curvature of the surface of the lens.

Answer 1

The light retraced its path if it is incident normally on a mirror. The ray after refraction through the lens and the liquid are parallel. We will apply the general thin lens equation with parameters.

n1=1, n2=1.5, n3=1.33, u=−0.2 m, and υ=∞

$\frac{n_3}{v} - \frac{n_1}{u} = \frac{n_2-n_1}{R} - \frac {n_3-n_2}{R}$

On solving for R, we get R=6.6 cm

Answer

6.6 cm