Question

A thin biconvex lens having refractive index 1. 5 is dipped in water having refractive index 1.33. It will behave like a .

• A. convergent lens
• B. divergent lens
• C. plane mirror

Answer 1

The correct option is A convergent lens

Given:
Refractive index of medium, ${\mu }_1=1.33$
Refractive index of lens, ${\mu }_2=1.5$

Let radius of curvature of the first surface be $R$.
Then, the radius of curvature of the second surface will be $-R$.

Using Lens Maker's Formula;
$\frac{1}{f}=(\frac{{\mu }_2}{{\mu }_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})$
$\frac{1}{f}=(\frac{1.5}{1.3}-1)(\frac{1}{R}+\frac{1}{R})$
$⟹f>0$
Since f is positive, parallel rays will form image to the right of the lens. Hence, it will behave as a convegent lens.