Question

Two thin symmetrical lenses of different nature and of different material have equal radii of curvature $R=15cm$. 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
$\left({\mu }_1-{\mu }_2\right)=\frac{4\times 15}{3\times 60}=\frac{1}{3}$