Question

A thin biconvex converging lens of refractive index $1.5$ has power of $+5\text{D}$. When this lens is immersed in a liquid, it acts as a diverging lens of focal length $100\text{cm}$. The refractive index of the liquid is

• A. $\frac{2}{3}$
• B. $\frac{15}{11}$
• C. $\frac{4}{3}$
• D. $\frac{5}{3}$

Answer 1

The correct option is B $\frac{15}{11}$

Given, for convergent lens,
refractive index of air, ${\mu }_a=1$,
refractive index of glass, ${\mu }_g=1.5$
Power, $P=+5D$.

For the radius of curvature, $R$ the power of the lens from lens makers formula can be written as,

$P=\frac{1}{f}=\left(\frac{{\mu }_g-{\mu }_a}{{\mu }_a}\right)\frac{2}{R}$

Substituting the values in above equation, we have,

$5=\left(\frac{1.5-1}{1}\right)\frac{2}{R}$

$\therefore R=20\text{cm}$

Let, the lens is immerged in the liquid of refractive index ${\mu }_l$ and the focal length of this lens in the medium is $f_m$.

Applying lens formula,

$\frac{1}{f_m}=\left(\frac{{\mu }_g-{\mu }_l}{{\mu }_l}\right)\frac{2}{R}$

Substituting the values in the above equation, we have

$\frac{1}{100}=(\frac{1.5}{{\mu }_l}-1)\frac{2}{20}$

$\Rightarrow \frac{1}{100}=(\frac{1.5}{{\mu }_l}-1)\frac{1}{10}$

$\Rightarrow {\mu }_l=\frac{15}{11}$

Hence, option (b) is correct answer.