Question

A convex lens made of a material of refractive index $1.5$ is placed in a medium having refractive index $1.6$, then it will behave as

• A. converging lens
• B. diverging lens
• C. plane glass plate
• D. Information insufficient

Answer 1

The correct option is B

diverging lens

Step 1: Given data

Given data :

The Refractive index of a convex lens in air is ${\mu }_g=1.5$

The Refractive index of a medium is ${\mu }_m=1.6$

Step 2: Finding the behavior of the lens

The formula of the focal length of the lens in air is $\frac{1}{f_a}=\left({}_a\mu _g-1\right)\left[\frac{1}{{\mu }_g}-\frac{1}{{\mu }_m}\right]$

We get the focal length of the lens in the other medium is,

$\frac{1}{f_m}=\left({}_m\mu _g-1\right)\left[\frac{1}{{\mu }_g}-\frac{1}{{\mu }_m}\right]=\frac{1}{f_a}\frac{\left({}_m\mu _g-1\right)}{\left({}_a\mu _g-1\right)}$

$\Rightarrow$ $\frac{1}{f_m}=\frac{1}{f_a}\left[\frac{{\displaystyle \frac{1.5}{1.6}}-1}{1.5-1}\right]$

$\frac{1}{f_m}=\frac{-0.125}{f_a}$

$\Rightarrow$ $f_m=\frac{-f_a}{0.125}$

The negative sign shows that the lens is diverging.

Hence, option B) is correct.