Question

The focal length of a convex lens immersed in a liquid of refractive index 1.2 is \( f_1 \) and its focal length when immersed in another liquid of refractive index 1.25 is \( f_2 \). If the refractive index of the material of the lens is 1.5, then \( f_1 : f_2 = \)

• A. 6 : 5
• B. 4 : 5
• C. 3 : 5
• D. 2 : 5

Hint:

For varying media, use the lens maker's formula and focus on the ratio \(\left(\frac{n_\text{lens}}{n_\text{medium}} - 1\right)\) to compare focal lengths.

Answer 1

The Correct Option is B

Focal length of lens in a medium is given by the lens maker’s formula: \[ \frac{1}{f} = \left( \frac{n_\text{lens}}{n_\text{medium}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] Thus, focal length \( f \propto \left( \frac{n_\text{lens}}{n_\text{medium}} - 1 \right)^{-1} \) \[ f_1 \propto \left( \frac{1.5}{1.2} - 1 \right)^{-1} = \left(0.25\right)^{-1} = 4 \] \[ f_2 \propto \left( \frac{1.5}{1.25} - 1 \right)^{-1} = \left(0.2\right)^{-1} = 5 \] \[ \therefore f_1 : f_2 = 4 : 5 \]