Question

A convex lens of focal length $40cm$ and refractive index $\frac{3}{2}$ is placed in a medium of refractive index $\frac{5}{3}$, then

• A. It will act as a convergent lens of focal length 200cm
• B. It will act as a convergent lens of focal length 100cm
• C. It will act as a divergent lens of focal length 100 cm
• D. It will act as a divergent lens of focal length 200 cm

Answer 1

Answer: (D)

It will act as a divergent lens of focal length 200 cm

LENS MAKERS FORMULA :

When a lens is immersed in a liquid: $\frac{1}{f^1}=(\frac{{\mu }_g}{{\mu }_l}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

Where ${\mu }_g$ = refractive index of lens

$f$ = focal length of lens in air

$f^1$ = focal length of lens in liquid

The new focal length $f^1$ is given by:

$\frac{f^1}{f}=\frac{({\mu }_g-1)}{(\frac{{\mu }_g}{{\mu }_l}-1)}$;

Given: ${\mu }_l=\frac{5}{3},{\mu }_g=\frac{3}{2}$

Substituting in the formula mentioned in the concept,

$\frac{f^1}{f}=\frac{\frac{3}{2}-1}{\frac{3\times 3}{2\times 5}-1}=-5$

$f^1=-5\times f=-5\times 40=-200cm$

As the focal length is negative, lens will be divergent.