Question

A convex lens of focal length 30 cm. is made of a material of refractive index $\frac{5}{3}$. When it is placed in a liquid its focal length is increased by 50 cm. The refractive index of the liquid is

• A. $\frac{5}{4}$
• B. $\frac{4}{3}$
• C. $\frac{7}{4}$
• D. $\frac{6}{5}$

Answer 1

Answer: (B)

$f_1=30cm;{\mu }_1=\frac{5}{3};{\mu }_2=?;f^{\prime }=80cm$

$\frac{f_1}{f_2}=\frac{({\mu }_1/{\mu }_2-1)}{({\mu }_1-1)}$

$\frac{30}{80}=\frac{(\frac{5}{3\mu }-1)}{\frac{2}{3}}$

$\frac{30}{80}\times \frac{2}{3}=(\frac{5}{3\mu }-1)$

$\frac{5}{4}=\frac{5}{3\mu }$

$\Rightarrow \mu =\frac{4}{3}$