An equiconvex glass lens (refractive index 1.5) has a focal length of 60 cm when it is placed in water (refractive index 4/3). The radius of curvature of each face is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 15 cm
When a lens is placed in water, the lens maker’s formula uses the refractive index of glass with respect to water. This relative index is \(n_{21}=n_g/n_w\). For an equiconvex lens, \(R_1=+R\) and \(R_2=-R\).
\( f=+60\ \mathrm{cm} \)
\( \begin{aligned} n_{21} &= \frac{n_g}{n_w} \\ &= \frac{1.5}{4/3} \\ &= \frac{9}{8} \end{aligned} \)
\( \begin{aligned} \frac{1}{f} &= (n_{21}-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ &= \left(\frac{9}{8}-1\right)\left(\frac{2}{R}\right) \\ &= \frac{1}{4R} \end{aligned} \)
\( \begin{aligned} \frac{1}{60} &= \frac{1}{4R} \\ R &= 15\ \mathrm{cm} \end{aligned} \)
Therefore, the radius of curvature of each face is 15 cm.
NCERT (2026-27):
Example 9.7 – “(iii) … For the same glass lens in water, n2 = 1.5, n1 = 1.33. Therefore,”
\( \displaystyle \frac{1.33}{f}=(1.5-1.33)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \)\( \qquad (9.26) \)