Q119 / 150 · MHT CET 2026 (16 April Shift 2) · Physics
A convex lens having refractive index 1.6 has focal length 12 cm, when it is in air. The focal length of that lens when placed in water is (refractive index of water = 1.28)
- A$655\ \text{mm}$
- Correct answerB$288\ \text{mm}$
- C$555\ \text{mm}$
- D$355\ \text{mm}$
Solution
Using the Lens Maker's formula for the lens in air:
$\dfrac{1}{f_a} = (\mu_g - 1) \left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right)$
$\dfrac{1}{12} = (1.6 - 1) \left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right)$
$\dfrac{1}{12} = 0.6 \left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right)$
Using the Lens Maker's formula for the lens in water:
$\dfrac{1}{f_w} = \left( \dfrac{\mu_g}{\mu_w} - 1 \right) \left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right)$
$\dfrac{1}{f_w} = \left( \dfrac{1.6}{1.28} - 1 \right) \left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right)$
$\dfrac{1}{f_w} = (1.25 - 1) \left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right)$
$\dfrac{1}{f_w} = 0.25 \left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right)$
Dividing the two equations:
$\dfrac{f_w}{f_a} = \dfrac{0.6}{0.25}$
$f_w = 12 \times 2.4 = 28.8\ \text{cm}$
$f_w = 288\ \text{mm}$
Answer: $288\ \text{mm}$
$\dfrac{1}{f_a} = (\mu_g - 1) \left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right)$
$\dfrac{1}{12} = (1.6 - 1) \left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right)$
$\dfrac{1}{12} = 0.6 \left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right)$
Using the Lens Maker's formula for the lens in water:
$\dfrac{1}{f_w} = \left( \dfrac{\mu_g}{\mu_w} - 1 \right) \left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right)$
$\dfrac{1}{f_w} = \left( \dfrac{1.6}{1.28} - 1 \right) \left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right)$
$\dfrac{1}{f_w} = (1.25 - 1) \left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right)$
$\dfrac{1}{f_w} = 0.25 \left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right)$
Dividing the two equations:
$\dfrac{f_w}{f_a} = \dfrac{0.6}{0.25}$
$f_w = 12 \times 2.4 = 28.8\ \text{cm}$
$f_w = 288\ \text{mm}$
Answer: $288\ \text{mm}$
Hence, the correct answer is option: 'B'.