Question

The radius of curvature of each surface of a convex lens having refractive index 1.8 is 20 cm. The lens is now immersed in a liquid of refractive index 1.5. The ratio of power of lens in air to its power in the liquid will be x : 1. The value of x is- [JEE MAIN-2023]

Answer 1

4

Answer 2

A symmetric biconvex lens (each surface radius R = 20 cm) in air obeys the lens‐maker’s formula

$\displaystyle \frac{1}{f_{\rm air}}=(n-1)\Bigl(\frac{1}{R_1}-\frac{1}{R_2}\Bigr)$.

For a symmetric lens, take $R_1=20$ cm and $R_2=-20$ cm so that

$\displaystyle \frac{1}{f_{\rm air}}=(1.8-1)\Bigl(\frac{1}{20}-\bigl(-\frac{1}{20}\bigr)\Bigr)=0.8\Bigl(\frac{2}{20}\Bigr)=0.08,$

i.e. $f_{\rm air}=12.5$ cm.

When immersed in a liquid of refractive index 1.5, the formula becomes

$\displaystyle \frac{1}{f_{\rm liq}}=\Bigl(\frac{n_{\rm lens}}{n_{\rm liq}}-1\Bigr)\Bigl(\frac{1}{R_1}-\frac{1}{R_2}\Bigr).$

Here,

$\displaystyle \frac{n_{\rm lens}}{n_{\rm liq}}-1=\frac{1.8}{1.5}-1=1.2-1=0.2,$

so

$\displaystyle \frac{1}{f_{\rm liq}}=0.2\Bigl(\frac{2}{20}\Bigr)=0.02,\quad f_{\rm liq}=50$ cm.

The power $P=1/f$ (with f in meters). Hence the ratio

$\displaystyle \frac{P_{\rm air}}{P_{\rm liq}}=\frac{f_{\rm liq}}{f_{\rm air}}=\frac{50}{12.5}=4.$

Thus, the ratio is $4:1$ and $x=4$.