Question

A convex lens of focal length 0.15 m is made of a material of refractive index 3/2. When it is placed in a liquid, its focal length is increased by 0.225 m. The refractive index of the liquid is

• A. $\frac{7}{4}$
• B. $\frac{5}{4}$
• C. $\frac{9}{4}$
• D. $\frac{3}{2}$

Answer 1

Answer: (B)

According to lens maker’s formula.
When the lens is in air, then
$\frac{1}{f_{air}}=\left({}^a{\mu }_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$\frac{1}{0.15}=\left(\frac{3}{2}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$\frac{1}{0.15}=\left(\frac{1}{2}\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ ...(i)
When the same lens is placed in a liquid of refractive index $\mu$, then
$\frac{1}{f_{liquid}}=\left(\frac{{}^a{\mu }_g}{{}^a{\mu }_l}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$\frac{1}{0.375}=\left(\frac{3/2}{\mu }-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$\frac{1}{0.375}=\left(\frac{3-2\mu }{2\mu }\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ ...(ii)
Dividing (i) by (ii), we get
$\frac{0.375}{0.15}=\frac{\mu }{3-2\mu }$
$0.375(3-3\mu )=0.15\mu$
$0.375\times 3=0.9\mu$
$\mu =\frac{0.375\times 3}{0.9}=\frac{5}{4}$