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 answer is –160 cm.

To find the focal length of the lens when submerged in a medium, we can use the Lens Maker's Formula. The Lens Maker's Formula relates the focal length ($f$) of a lens to the refractive index of its material (${\mu }_l$), the refractive index of the surrounding medium (${\mu }_m$), and the radii of curvature of its two surfaces ($R_1$ and $R_2$):

$\frac{1}{f}=\left(\frac{{\mu }_l}{{\mu }_m}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Step 1: Write down the expression for the lens in air
When the lens is in air, the refractive index of the surrounding medium is ${\mu }_m=1$.
Given:
- Focal length in air ($f_a$) = 20 cm
- Refractive index of the lens (${\mu }_l$) = 1.5

Substituting these values into the Lens Maker's Formula:
$\frac{1}{20}=\left(\frac{1.5}{1}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$\frac{1}{20}=0.5\cdot \left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\frac{1}{20\cdot 0.5}=\frac{1}{10}$

Step 2: Write down the expression for the lens submerged in the medium
When the lens is submerged in the medium of refractive index ${\mu }_m=1.6$, let its focal length be $f_m$.

Using the formula again:
$\frac{1}{f_m}=\left(\frac{1.5}{1.6}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Step 3: Calculate the new focal length
Substitute the value of $\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\frac{1}{10}$ into the equation:
$\frac{1}{f_m}=\left(\frac{1.5}{1.6}-1\right)\cdot \frac{1}{10}$
$\frac{1}{f_m}=\left(\frac{15}{16}-1\right)\cdot \frac{1}{10}$
$\frac{1}{f_m}=\left(-\frac{1}{16}\right)\cdot \frac{1}{10}$
$\frac{1}{f_m}=-\frac{1}{160}$
$f_m=-160\text{cm}$

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