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 given by the formula: \[ \frac{n_2
- n_1}{f} = \frac{n_2}{R} \] where:
- \( n_2 = n \) (refractive index of the medium),
- \( n_1 = 1 \) (refractive index of air),
- \( R \) is the radius of curvature of the convex surface. Substituting the values: \[ \frac{n
- 1}{f} = \frac{n}{R} \] \[ f = \frac{R}{n
- 1} \] Now, using the mirror equation: \[ \frac{1}{f} = \frac{1}{v}
- \frac{1}{u} \] Where:
- \( u =
-\frac{R}{3} \) (object distance),
- \( v \) is the image distance (which we need to find). Substituting \( u =
-\frac{R}{3} \) and \( f = \frac{R}{n
- 1} \): \[ \frac{1}{\frac{R}{n
- 1}} = \frac{1}{v} + \frac{3}{R} \] Solving for \( v \): \[ v = \frac{R(n
- 1)}{3n
- 4} \] Thus, the position of the image is given by: \[ v = \frac{R(n
- 1)}{3n
- 4} \] This gives the image position based on the refractive index \( n \) and the radius of curvature \( R \).