A concave lens of glass of refractive index 1.5 has both surfaces of the same radius of curvature R. On immersing the lens in a medium of refractive index 1.75, it will behave as a
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → converging lens of focal length 3.5 R
In a medium, the lens maker’s formula uses the refractive index of the glass relative to the surrounding medium. For a biconcave lens (Cartesian sign convention, light from the left) \( R_1 = -R \) and \( R_2 = +R \).
\( \displaystyle \frac{1}{f} = \left(\frac{\mu_g}{\mu_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)
\( \displaystyle \frac{\mu_g}{\mu_m} - 1 = \frac{1.5}{1.75} - 1 = \frac{6}{7} - 1 = -\frac{1}{7} \)
\( \displaystyle \frac{1}{R_1} - \frac{1}{R_2} = -\frac{1}{R} - \frac{1}{R} = -\frac{2}{R} \)
\( \displaystyle \frac{1}{f} = \left(-\frac{1}{7}\right)\left(-\frac{2}{R}\right) = \frac{2}{7R} \)
\( \displaystyle f = +3.5\,R \)
The focal length is positive, so the lens now converges light. It happens because the glass is optically rarer than the surrounding medium, which reverses the action of the lens.