Question

Two thin symmetrical lenses of different nature and of different material have equal radii of curvature $R=15 \, cm$. The lenses are put close together and immersed in water $\left(\mu_{w}=\frac{4}{3}\right)$. The focal length of the system in water is 30 $cm$. The difference between refractive of the two lenses is

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{3}$
• D. $\frac{3}{4}$

Answer 1

Answer: (C)

Let $f_{1}$ and $f_{2}$ be the focal length in water. Then
$\frac{1}{f_{1}}=\left(\frac{\mu_{1}}{\mu_{w}}-1\right)\left(\frac{1}{R}+\frac{1}{R}\right)=\left(\frac{\mu_{1}}{\mu_{w}}-1\right)\left(\frac{2}{R}\right) ....$(1)
$\frac{1}{f_{1}}=\left(\frac{\mu_{2}}{\mu_{w}}-1\right)\left(-\frac{1}{R}+\frac{1}{R}\right)=\left(\frac{\mu_{2}}{\mu_{w}}-1\right)\left(-\frac{2}{R}\right)....$(2)
Adding (1) and (2), we get
$\frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{2\left(\mu_{1}-\mu_{2}\right)}{\mu_{w} R}$
or $\frac{1}{30}=\frac{2\left(\mu_{1}-\mu_{2}\right)}{\mu_{w} R}$
$\therefore \left(\mu_{1}-\mu_{2}\right)=\frac{\mu_{w} R}{60}$
Substituting the values
$\left(\mu_{1}-\mu_{2}\right)=\frac{4 \times 15}{3 \times 60}=\frac{1}{3}$