A diverging lens of glass \( (\mu = 1.5) \) has both surfaces of the same radius of curvature \( (R) \). On immersion in a medium of refractive index \( (\mu = 1.8) \), it will behave as
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → converging lens with focal length \( f = 3R \)
The lens maker's formula for a lens of index \( \mu_{l} \) in a medium of index \( \mu_{m} \) is
\( \displaystyle \frac{1}{f} = \left(\frac{\mu_{l}}{\mu_{m}} - 1\right)\left(\frac{1}{R_{1}} - \frac{1}{R_{2}}\right) \)
A diverging lens with equal radii is biconcave: with the sign convention, \( R_{1} = -R \) and \( R_{2} = +R \), so \( \frac{1}{R_{1}} - \frac{1}{R_{2}} = -\frac{2}{R} \).
\( \displaystyle \frac{\mu_{l}}{\mu_{m}} - 1 = \frac{1.5}{1.8} - 1 = -\frac{1}{6} \)
\( \displaystyle \frac{1}{f} = \left(-\frac{1}{6}\right)\left(-\frac{2}{R}\right) = \frac{1}{3R} \;\Rightarrow\; f = +3R \)
The focal length is positive, so the lens now converges: its index is lower than that of the surrounding medium, which reverses its behaviour. The focal length is \( 3R \).