Question

A thin equiconvex lens is made of a glass of refractive index 1.5 and its focal length is 0.2 m. If it acts as a concave lens of 0.5 m focal length when dipped in a liquid, the refractive index of the liquid is:

• A. $\frac{17}{8}$
• B. $\frac{15}{8}$
• C. $\frac{13}{8}$
• D. $\frac{9}{8}$

Answer 1

Answer: (B)

Given,

${\mu }_1=1.5f_1=0.2m{f}_2=(-0.5){\mu }_2=?$

Using lens makers formula and taking ratios, $\frac{f_1}{f_2}=\frac{({\mu }_2-1)}{({\mu }_1-1)}$

$\Rightarrow \frac{0.2}{(-0.5)}=\frac{(\frac{1.5}{{\mu }_e}-1)}{(1.5-1)}$

$\Rightarrow \frac{0.2\times 0.5}{0.5}=(1-\frac{1.5}{{\mu }_e})$

$\Rightarrow \frac{1.5}{{\mu }_e}=0.8$

$\Rightarrow {\mu }_e=\frac{15}{8}$