A convex lens has 20 cm focal length in air. Its focal length in water would be
(Given: Refractive index of water = 1.33, refractive index for glass = 1.5)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → +78.2 cm
By the lens maker's formula, the power of a lens depends on the refractive index of the glass relative to the surrounding medium; the radii of curvature do not change when the lens is put in water.
\( \displaystyle \frac{1}{f} = \left(\frac{\mu_{g}}{\mu_{m}} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)
Dividing the equation for air by the equation for water:
\( \displaystyle \frac{f_{w}}{f_{a}} = \frac{\mu_{g} - 1}{(\mu_{g}/\mu_{w}) - 1} \)
\( \mu_{g} - 1 = 1.5 - 1 = 0.5 \)
\( \mu_{g}/\mu_{w} - 1 = 1.5/1.33 - 1 = 1.128 - 1 = 0.128 \)
\( \displaystyle f_{w} = 20\,\mathrm{cm} \times \frac{0.5}{0.128} = 78.2\,\mathrm{cm} \)
The glass is still optically denser than water, so the lens remains converging and the focal length is positive: \( +78.2\,\mathrm{cm} \).
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Power of a Lens – “… (iii) A convex lens has 20 cm focal length in air. What is focal length in water? (Refractive index of air-water = 1.33, refractive index for air-glass = 1.5.)”
Power of a Lens – “… (iii) For a glass lens in air, n2 = 1.5, n1 = 1, f = +20 cm. Hence, the lens formula gives
\( \displaystyle \frac{1}{20} = 0.5\left[\frac{1}{R_1} - \frac{1}{R_2}\right] \)
For the same glass lens in water, n2 = 1.5, n1 = 1.33. Therefore,
\( \displaystyle \frac{1.33}{f} = (1.5 - 1.33)\left[\frac{1}{R_1} - \frac{1}{R_2}\right] \qquad (9.26) \)
Combining these two equations, we find f = + 78.2 cm.”