Question

Object is placed at $40 \text{ cm}$ from spherical surface whose radius of curvature is $20 \text{ cm}$. Find height of image formed.

• A. $2 \text{ cm}$
• B. $4 \text{ cm}$
• C. $0.96 \text{ cm}$
• D. $1.96 \text{ cm}$

Answer 1

The Correct Option is C

Step 1: Use the formula for refraction at a spherical surface.
$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$.
Given: $\mu_1 = 1$, $\mu_2 = 1.54$, $u = -40 \text{ cm}$, $R = -20 \text{ cm}$.

Step 2: Calculate the image distance $v$.
$\frac{1.54}{v} - \frac{1}{-40} = \frac{1.54 - 1}{-20} \implies \frac{1.54}{v} + \frac{1}{40} = \frac{0.54}{-20}$.
$\frac{1.54}{v} = -0.027 - 0.025 = -0.052$.
$v = \frac{1.54}{-0.052} \approx -29.61 \text{ cm}$.

Step 3: Use the magnification formula for spherical refraction.
$m = \frac{h_i}{h_o} = \frac{\mu_1 v}{\mu_2 u}$.

Step 4: Calculate image height $h_i$.
$\frac{h_i}{2} = \frac{1 \times (-29.61)}{1.54 \times (-40)} \implies \frac{h_i}{2} = \frac{-29.61}{-61.6} \approx 0.48$.
$h_i = 0.48 \times 2 = 0.96 \text{ cm}$.

Final Answer: $0.96 \text{ cm}$. Correct option is (3).