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
$\frac{0.375}{0.15} = \frac{\mu}{3 - 2\mu}$
$0.375 (3-3\mu)=0.15\mu$
$0.375 \times 3 =0.9 \mu$
$\mu = \frac{0.375 \times 3}{0.9} = \frac{5}{4}$