Question

A point source, in air, is placed at a distance of 6 cm in front of a convex spherical surface (n = 1.5 and radius of curvature = 24 cm). Find the position and nature of the image formed.

Hint:

Always follow the Cartesian sign convention: distances measured in the direction of incident light are positive; opposite are negative.

Answer 1

Step 1: Understanding the Concept:
When light travels from one medium to another through a spherical interface, the image formation is governed by the refraction formula for spherical surfaces.

Step 2: Key Formula or Approach:
The refraction formula is: \[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \]

Step 3: Detailed Explanation:
Given: - \(n_1 = 1\) (air) - \(n_2 = 1.5\) (glass/medium) - \(u = -6\) cm (object distance, sign convention) - \(R = +24\) cm (convex surface) Substituting into the formula: \[ \frac{1.5}{v} - \frac{1}{-6} = \frac{1.5 - 1}{24} \] \[ \frac{1.5}{v} + \frac{1}{6} = \frac{0.5}{24} \] \[ \frac{1.5}{v} = \frac{1}{48} - \frac{1}{6} \] \[ \frac{1.5}{v} = \frac{1 - 8}{48} = -\frac{7}{48} \] \[ v = -\frac{1.5 \times 48}{7} = -\frac{72}{7} \approx -10.29 \text{ cm} \]

Step 4: Final Answer:
The image is formed at a distance of approximately 10.29 cm from the pole on the same side as the object. Since \(v\) is negative, the image is virtual.