A thin equiconvex lens of radius of curvature R made up of glass (\( \mu = 1.5 \)) is dipped in water (\( \mu = 1.3 \)). Its focal length in water is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 3.25\,R \)
The lens maker's formula for a lens of refractive index \( \mu_g \) in a medium of refractive index \( \mu_m \) is
\( \displaystyle \frac{1}{f} = \left(\frac{\mu_g}{\mu_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)
For an equiconvex lens \( R_1 = +R \) and \( R_2 = -R \), so \( 1/R_1 - 1/R_2 = 2/R \).
\( \displaystyle \frac{1}{f} = \left(\frac{1.5}{1.3} - 1\right)\frac{2}{R} = \frac{0.2}{1.3} \times \frac{2}{R} = \frac{0.4}{1.3\,R} \)
\( \displaystyle f = \frac{1.3}{0.4}\,R = 3.25\,R \)