Question

A convex lens made of glass of refractive index 1.5 is immersed in a liquid. If the focal length of the lens when immersed in the liquid is twice its focal length when it is in air, then the refractive index of the liquid is

• A. 1.6
• B. 1.4
• C. 1.2
• D. 1.3

Hint:

If a lens is immersed in a medium with refractive index closer to its own, its focal length increases (power decreases).

Answer 1

The Correct Option is C

Step 1: Lens Maker's Formula:
$\frac{1}{f} = (\mu_{rel} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$. Let $K = \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$. In air ($\mu_{air} \approx 1$): \[ \frac{1}{f_a} = (\mu_g - 1)K = (1.5 - 1)K = 0.5K \] In liquid ($\mu_l$): \[ \frac{1}{f_l} = \left( \frac{\mu_g}{\mu_l} - 1 \right)K = \left( \frac{1.5}{\mu_l} - 1 \right)K \]
Step 2: Apply Given Condition:
Given $f_l = 2 f_a$. \[ \frac{1}{f_l} = \frac{1}{2 f_a} \] Substitute expressions: \[ \left( \frac{1.5}{\mu_l} - 1 \right)K = \frac{1}{2} (0.5K) \] \[ \frac{1.5}{\mu_l} - 1 = 0.25 \] \[ \frac{1.5}{\mu_l} = 1.25 \] \[ \mu_l = \frac{1.5}{1.25} = \frac{150}{125} = \frac{6}{5} \] \[ \mu_l = 1.2 \]
Step 4: Final Answer:
The refractive index of the liquid is 1.2.