Question

A convex lens of focal length 0.2 m and made of glass ${(}^a{\mu }_g=1.5)$ is immersed in water ${(}^a{\mu }_w=1.33)$. Find the change in the focal length of the lens.

• A. 5.8 m
• B. 158 m
• C. 0.58 m
• D. 5.8 cm

Answer 1

Answer: (C)

0.58 m

We know ,

From Lens maker formula:

$\frac{1}{f}=({\mu }_g-{\mu }_m)(\frac{1}{R_1}-\frac{1}{R_2})$

where,

$f=$ focal length

${\mu }_g=$ refractive index of glass $-1.5$

${\mu }_m$ - refractive index of medium w.r.t air

$R_1$ and $R_2$ are radius of curvature

Refractive index of air is 1

Hence focal length of lens in air

$\frac{1}{f_a}=(1.5-1)(\frac{1}{R_1}-\frac{1}{R_2})$

Refractive index of water = 1.33

Hence focal length of water

$\frac{1}{f_w}=(1.5-1.33)(\frac{1}{R_1}-\frac{1}{R_2})$

From above two equation we get

$\frac{f_w}{f_a}=\frac{1.5-1}{1.5-1.33}$

$⟹\frac{f_w}{f_a}=\frac{0.5}{0.17}=2.94$

GIven, Focal length of lens in air, $f_a=0.2m$

$\therefore$ Focal length in water , $f_w=0.2\times 2.94=0.58m$