Question

A point source of light in air is kept at a distance of 12 cm in front of a convex spherical surface of glass of refractive index 1.5 and radius of curvature 30 cm. Find the nature and position of the image formed.

Hint:

Use sign conventions carefully: object distances are negative if measured toward the surface; radius is positive for convex surfaces. A negative image distance implies a virtual image on the same side as the object.

Answer 1

Applying the formula for refraction at a spherical surface:\[\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}\]Given:- \( \mu_1 = 1 \) (air), \( \mu_2 = 1.5 \) (glass)- \( u = -12\ \text{cm} \) (object distance)- \( R = +30\ \text{cm} \) (radius of curvature for a convex surface)Substitution yields:\[\frac{1.5}{v} - \frac{1}{-12} = \frac{1.5 - 1}{30} = \frac{0.5}{30}\Rightarrow \frac{1.5}{v} + \frac{1}{12} = \frac{1}{60}\Rightarrow \frac{1.5}{v} = \frac{1}{60} - \frac{1}{12} = \frac{1 - 5}{60} = -\frac{4}{60}\Rightarrow \frac{1.5}{v} = -\frac{1}{15}\Rightarrow v = -22.5\ \text{cm}\]Therefore, the image is located 22.5 cm from the surface, on the same side as the object, and is virtual.