The focal length of a thin lens in vacuum is \( f \). If the material of the lens has refractive index \( \displaystyle\frac{3}{2} \), its focal length when immersed in water of refractive index \( \displaystyle\frac{4}{3} \) would be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 4f \)
By the lens maker’s formula, the focal length of a thin lens in a medium depends on the refractive index of the lens material relative to the medium, \( \displaystyle\frac{1}{f} = \left(\frac{\mu_g}{\mu_m}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \). The radii of curvature do not change, so the bracket with the radii is a common constant K.
In vacuum (\( \mu_m = 1 \))
\( \displaystyle \frac{1}{f} = \left(\frac{3}{2}-1\right)K = \frac{K}{2} \)
In water (\( \mu_m = 4/3 \))
\( \displaystyle \frac{\mu_g}{\mu_m} = \frac{3/2}{4/3} = \frac{9}{8} \)
\( \displaystyle \frac{1}{f'} = \left(\frac{9}{8}-1\right)K = \frac{K}{8} \)
Dividing the two results:
\( \displaystyle \frac{f'}{f} = \frac{K/2}{K/8} = 4 \Rightarrow f' = 4f \)
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. 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.)”