Question

A convex lens of focal length 20 cm made of glass of refractive index 1.5 is immersed in water having refractive index 1.33. The change in the focal length of lens is

• A. 62.2 cm
• B. 5.82 cm
• C. 58.2 cm
• D. 6.22 cm

Answer 1

Answer: (C)

When the lens is in air
$\frac{1}{f_a}=\left({\mu }_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$\frac{1}{20}=\left(1.5-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ ...(i)
When lens is in water,
$\frac{1}{f_w}=\left(\frac{{\mu }_g}{{\mu }_w}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
or $\frac{1}{f_w}=\left(\frac{1.5-1.33}{1.33}\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ ....(ii)
Divide (i) by (ii), we get
or, $\frac{f_w}{20}=\left(1.5-1\right)\left(\frac{1.33}{1.5-1.33}\right)$
or, $f_w=20\times 0.5\times \frac{1.33}{0.17}=78.2cm$
The change in focal length = $78.2-20=58.2cm$