If a biconvex lens of material of refractive index 1.5 has focal length 20 cm in air, then its focal length when it is submerged in a medium of refractive index 1.6 is

The correct option is –160 cm.

To find the focal length of the biconvex lens when submerged in the medium, we can use the lens maker's formula.
The lens maker's formula in general form is:
$\frac{1}{f}=\left(\frac{{\mu }_{\mathrm{lens}}}{{\mu }_{\mathrm{medium}}}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Let the refractive index of the lens material be ${\mu }_g=1.5$.
Let $R_1$ and $R_2$ be the radii of curvature of the two surfaces of the biconvex lens.

Step 1: Lens in air
When the lens is in air, the refractive index of the surrounding medium is ${\mu }_{\mathrm{air}}=1.0$, and its focal length is $f_a=20\text{cm}$.
Applying the lens maker's formula:
$\frac{1}{f_a}=\left({\mu }_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
Substituting the given values:
$\frac{1}{20}=\left(1.5-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$\frac{1}{20}=0.5\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\frac{1}{20\times 0.5}=\frac{1}{10}$     --- (Equation 1)

Step 2: Lens submerged in the medium
When the lens is submerged in the medium of refractive index ${\mu }_m=1.6$, let its new focal length be $f_m$.
Applying the lens maker's formula again:
$\frac{1}{f_m}=\left(\frac{{\mu }_g}{{\mu }_m}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
Substituting the value of the curvature term from Equation 1:
$\frac{1}{f_m}=\left(\frac{1.5}{1.6}-1\right)\left(\frac{1}{10}\right)$
$\frac{1}{f_m}=\left(\frac{1.5-1.6}{1.6}\right)\left(\frac{1}{10}\right)$
$\frac{1}{f_m}=\left(\frac{-0.1}{1.6}\right)\left(\frac{1}{10}\right)$
$\frac{1}{f_m}=-\frac{1}{16}\times \frac{1}{10}$
$\frac{1}{f_m}=-\frac{1}{160}$
$f_m=-160\text{cm}$

Thus, the focal length of the lens when submerged in the medium is –160 cm. The negative sign indicates that the lens behaves as a diverging lens in this medium because the surrounding medium is optically denser than the lens material.