Question

A transparent sphere of refractive index ' \(\mu\) ' and radius of curvature ' \(R\) ' is kept in air. A point object is placed at a distance ' \(d\) ' from the surface of the sphere so that the real image is formed at the same distance ' \(d\) ' from exactly opposite side of the sphere. The distance ' \(d\) ' is

• A. \(\frac{\mu}{R}\)
• B. \(R(\mu - 1)\)
• C. \(\frac{R}{(\mu - 1)}\)
• D. \(\frac{R}{(\mu + 1)}\)

Hint:

Symmetric refraction: Object distance = image distance $\Rightarrow$ special condition

Answer 1

The Correct Option is C

Concept: Refraction at spherical surface: \[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \]

Step 1: First refraction (air to sphere). \[ \frac{\mu}{v_1} - \frac{1}{-d} = \frac{\mu - 1}{R} \]

Step 2: Second refraction (sphere to air). \[ \frac{1}{d} - \frac{\mu}{-v_1} = \frac{1 - \mu}{R} \]

Step 3: Solve symmetry condition.
Using image at same distance on other side gives: \[ d = \frac{R}{\mu - 1} \] Final Answer: Option (C)