Question

A convex lens of refractive index 1.5 has power 3D. It is placed in a liquid of refractive index 2. The new power of the lens is

• A. 3 D
• B. 0.75 D
• C. 1.5 D
• D. 2 D

Hint:

Power decreases when a lens is placed in a medium with refractive index $> 1$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find the change in power of a lens when it is immersed in a liquid with a different refractive index.
Step 2: Key Formula or Approach:
Lens Maker's Formula: \( P = \frac{1}{f} = (\mu_{\text{rel}} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \).
Step 3: Detailed Explanation:
In air: \( P_a = (1.5 - 1) K = 0.5 K = 3 \text{ D} \), where \( K \) represents the curvature part.
In liquid: \( P_l = \left( \frac{1.5}{2} - 1 \right) K = (0.75 - 1) K = -0.25 K \).
Taking the ratio:
\[ \frac{P_l}{P_a} = \frac{-0.25 K}{0.5 K} = -\frac{1}{2} \] \[ P_l = -\frac{1}{2} \times 3 \text{ D} = -1.5 \text{ D} \] The magnitude of the new power is 1.5 D. (The negative sign indicates it now behaves as a concave lens).
Step 4: Final Answer:
The new power of the lens is 1.5 D.