Question

A convex lens of focal length 20 cm is immersed in a liquid of refractive index 1.3. If the refractive index of the material of the lens is 1.5, then the focal length of the lens when immersed in the liquid is:

• A. \( 20 \) cm
• B. \( 35 \) cm
• C. \( \mathbf{65} \) cm
• D. \( 40 \) cm

Hint:

- The focal length of a lens in a medium is calculated using the modified lens maker’s formula. - The new focal length increases when the lens is placed in a medium with a refractive index closer to that of the lens. - Always substitute values carefully to avoid calculation mistakes.

Answer 1

The Correct Option is C

Step 1: Lens Maker's Formula in a Medium The focal length of a lens in a medium is given by the modified lens maker’s formula: \[ \frac{1}{f_m} = \left( \frac{n_{\text{lens}}}{n_{\text{medium}}} - 1 \right) \frac{1}{f} \] where: - \( f_m \) is the focal length in the medium, - \( n_{\text{lens}} = 1.5 \) is the refractive index of the lens, - \( n_{\text{medium}} = 1.3 \) is the refractive index of the medium, - \( f = 20 \) cm is the original focal length in air. 

Step 2: Substituting Given Values \[ \frac{1}{f_m} = \left( \frac{1.5}{1.3} - 1 \right) \frac{1}{20} \] \[ = \left( \frac{1.5 - 1.3}{1.3} \right) \frac{1}{20} \] \[ = \left( \frac{0.2}{1.3} \right) \frac{1}{20} \] 

Step 3: Solving for \( f_m \) \[ f_m = \frac{20 \times 1.3}{0.2} \] \[ f_m = \frac{26}{0.2} = 65 \text{ cm} \] 

Step 4: Verifying the Correct Option Comparing with the given options, the correct answer is: \[ \mathbf{65} \text{ cm} \]