Question

A point object is placed in air at a distance \( \frac{R}{3} \) in front of a convex surface of radius of curvature \( R \), separating air from a medium of refractive index \( n \) (where \( n<4 \)). Find the nature and position of the image formed.

Hint:

In problems involving spherical surfaces, always apply the mirror equation carefully, and remember to use the sign convention for distances. The focal length for a spherical surface separating two media can be derived using the lens maker's formula.

Answer 1

The focal length \( f \) of a spherical surface separating two media is described by the formula:\[\frac{n_2 - n_1}{f} = \frac{n_2}{R}\]With the following definitions:
- \( n_2 = n \) (refractive index of the medium),
- \( n_1 = 1 \) (refractive index of air),
- \( R \) representing the radius of curvature of the convex surface.Substituting these values yields:\[\frac{n - 1}{f} = \frac{n}{R}\]Which simplifies to:\[f = \frac{R}{n - 1}\]Utilizing the mirror equation:\[\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\]Where:
- \( u = -\frac{R}{3} \) (object distance),
- \( v \) is the image distance, the value to be determined.Substituting \( u = -\frac{R}{3} \) and \( f = \frac{R}{n - 1} \) into the mirror equation:\[\frac{1}{\frac{R}{n - 1}} = \frac{1}{v} + \frac{3}{R}\]The calculation for \( v \) results in:\[v = \frac{R(n - 1)}{3n - 4}\]Therefore, the image position is determined by:\[v = \frac{R(n - 1)}{3n - 4}\]This equation provides the image position as a function of the refractive index \( n \) and the radius of curvature \( R \).