Question

A concave lens (refractive index $= 1.5$) has both surfaces of same radius of curvature R . If it is immersed in a liquid of refractive index 1.75 it will act as a}

• A. concave lens of focal length (3.5) R
• B. concave lens of focal length 2 R .
• C. convex lens of focal length (3.5) R
• D. convex lens of focal length $2R$

Hint:

If $\mu_{medium}>\mu_{lens}$, the nature of the lens reverses (Concave $\to$ Convex).

Answer 1

The Correct Option is C

Step 1: Lens Maker's Formula
$\frac{1}{f} = (\frac{\mu_l}{\mu_m} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$.
For concave lens, $R_1 = -R, R_2 = +R$. So $(\frac{1}{R_1} - \frac{1}{R_2}) = -\frac{2}{R}$.
Step 2: Calculation
$\mu_l = 1.5$ (lens), $\mu_m = 1.75$ (liquid).
$\frac{1}{f'} = (\frac{1.5}{1.75} - 1)(-\frac{2}{R}) = (\frac{6}{7} - 1)(-\frac{2}{R}) = (-\frac{1}{7})(-\frac{2}{R}) = \frac{2}{7R}$.
Step 3: Conclusion
$f' = +3.5R$. Since focal length is positive, it acts as a **convex (converging)** lens.
Final Answer: (C)