Read the information given below.
It is useful to design lenses of derived focal length using surface of suitable radii of curvature. Lens maker formula helps in finding various unknowns and is widely applied.
A convex lens has 30 cm focal length in air. What is its focal length in a liquid whose refractive index is \( \displaystyle\frac{4}{3} \)? (Given refractive index of glass is \( \displaystyle\frac{3}{2} \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( +120\,\mathrm{cm} \)
By the lens maker's formula, the focal length depends on the relative refractive index of the glass with respect to the surrounding medium; the radii of curvature are the same in air and in the liquid.
\( \displaystyle \frac{1}{f_{\mathrm{air}}} = (\mu_g - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)
\( \displaystyle \frac{1}{f_{\mathrm{liq}}} = \left(\frac{\mu_g}{\mu_l} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)
Dividing the first by the second:
\( \displaystyle \frac{f_{\mathrm{liq}}}{f_{\mathrm{air}}} = \frac{\mu_g - 1}{\mu_g/\mu_l - 1} \)
\( \mu_g - 1 = 1.5 - 1 = 0.5 \)
\( \mu_g/\mu_l = 1.5 \times 3/4 = 1.125 \)
\( \displaystyle \frac{f_{\mathrm{liq}}}{f_{\mathrm{air}}} = \frac{0.5}{0.125} = 4 \)
\( \displaystyle f_{\mathrm{liq}} = 4 \times 30\,\mathrm{cm} = +120\,\mathrm{cm} \)
Since \( \mu_g \gt \mu_l \), the lens is still converging, so the focal length stays positive.
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.)”