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: Principle used.
For refraction at a spherical surface, the relation between object distance and image distance is: \[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \]
Step 2: Substitute given values.
Given:
\( n_1 = 1 \) (air)
\( n_2 = 1.5 \) (glass)
\( u = -6 \, \text{cm} \)
\( R = +24 \, \text{cm} \)
\[ \frac{1.5}{v} - \frac{1}{-6} = \frac{1.5 - 1}{24} \]
\[ \frac{1.5}{v} + \frac{1}{6} = \frac{0.5}{24} \]
\[ \frac{1.5}{v} = \frac{1}{48} - \frac{1}{6} \]
\[ \frac{1.5}{v} = \frac{1 - 8}{48} = -\frac{7}{48} \]
\[ v = -\frac{1.5 \times 48}{7} = -\frac{72}{7} \]
\[ v \approx -10.29 \, \text{cm} \]
Final Answer:
Image distance \( v \approx -10.29 \, \text{cm} \).
Since \( v \) is negative, the image is virtual and formed on the same side as the object.