Question

A lens has a power of $10\text{D}$ when placed in air. When it is immersed in water$(\mu =\frac{4}{3})$, the change in power is
(Refractive index of lens material is $1.5$)

• A. $2.55\text{D}$
• B. $5.0\text{D}$
• C. $6.5\text{D}$
• D. $-7.5\text{D}$

Answer 1

The correct option is D $-7.5\text{D}$

Given:
Power of lens $P=10\text{D}$

Refractive index of water$=\frac{4}{3}$

Refractive index of material $=\frac{3}{2}=1.5$

As we know that lens maker formula is given by,

$\frac{1}{f}=\left(\frac{{\mu }_2}{{\mu }_1}\right)(\frac{1}{R_1}-\frac{1}{R_2})$

and power is $P=\frac{1}{f}$

When the lens is in air then the power by the lens maker formula,

$10=(1.5-1)(\frac{1}{R_1}-\frac{1}{R_2})$

$(\frac{1}{R_1}-\frac{1}{R_2})=K=\frac{10}{0.5}=20{\text{m}}^{-1}$

Again, when the lens immersed in water by using the lens maker formula,

$P=(\frac{1.5}{\frac{4}{3}}-1)K=(\frac{1.5}{\frac{4}{3}}-1)20$

$P=\left(\frac{4.5-4}{4}\right)20=2.5\text{D}$

Change in power is,

$=2.5-10\text{D}$

$=-7.5\text{D}$