Question

A double concave lens is placed in water of refractive index $\frac{4}{3}$. If the radii of curvature of the two surfaces of the lens are 10 cm and 30 cm; and the refractive index of glass is $\frac{3}{2}$, find the focal length of the lens in water.

• A. 60 cm
• B. -60 cm
• C. 30 cm
• D. - 30 cm

Answer 1

The correct option is B -60 cm

$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$
$=(\frac{{\mu }_2}{{\mu }_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})$
Here ${\mu }_2=\frac{3}{2},{\mu }_1=\frac{4}{3},R_1=-10cm$ and $R_2=30cm$.
$\frac{1}{f}=(\frac{3}{2}\times \frac{3}{4}-1)(\frac{1}{-10}-\frac{1}{30})$
$=(\frac{9}{8}-1)\left(\frac{-40}{10\times 30}\right)$
$=-\frac{1}{8}\times \frac{40}{10\times 30}=-\frac{1}{60}$
$f=-60cm$