Question

A double convex lens $(\mu =1.5)$ in air has its focal length equal to $20cm$. When immersed in water $(\mu =4/3)$ its focal length will be

• A. $20cm$
• B. $80/9cm$
• C. $360/9cm$
• D. $80cm$

Answer 1

The correct option is D $80cm$

Given,

Refractive indexes for, glass ${\mu }_g=1.5$, water ${\mu }_w=\frac{4}{3}$ and ${\mu }_a=1$ .

If f is the focal length of the lens in air then

$\frac{1}{f_a}=(\frac{{\mu }_g}{{\mu }_a}-1)\times (\frac{1}{R_1}-\frac{1}{R_2}).........\left(1\right)$

If f is the focal length of the lens in water then
$\frac{1}{f_w}=(\frac{{\mu }_g}{{\mu }_w}-1)\times (\frac{1}{R_1}-\frac{1}{R_2}).........\left(2\right)$

Divide equation (1) by (2)

$\frac{\frac{1}{f_a}}{\frac{1}{f_w}}=\frac{(\frac{{\mu }_g}{{\mu }_a}-1)}{(\frac{{\mu }_g}{{\mu }_w}-1)}$

$f_w=(\frac{{\mu }_g-{\mu }_a}{{\mu }_a}\times \frac{{\mu }_w}{{\mu }_g-{\mu }_w})f_a$

$f_w=(\frac{1.5-1}{1}\times \frac{4/3}{1.5-4/3})\times 20=80cm$

Hence, focal length in water, $f_w=80cm$