Question

An equiconvex air lens $(R_1=R_2=10\text{cm})$ is made in glass medium having refractive index $(\mu =\frac{3}{2})$. Find the refractive index of the material to be filled in, so that the power of the lens changes without change in its magnitude.

• A. $\frac{3}{2}$
• B. $2$
• C. $1.8$
• D. $1.2$

Answer 1

The correct option is B $2$

Given,
Refractive index of glass = $\frac{3}{2}$
$R_1=R_2=10\text{cm}$

As we know that the lens maker formula is given by

$\frac{1}{f}=(\frac{{\mu }_2}{{\mu }_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

For Case$-I$

$\frac{1}{f}=(\frac{2}{3}-1)(\frac{1}{R}-\frac{1}{-R})$

$\frac{1}{f}=(\frac{2}{3}-1)\frac{2}{R}.......\left(1\right)$

For case$-II$

$\frac{1}{-f}=(\frac{2\mu }{3}-1)(\frac{1}{R}-\frac{1}{-R})$

$\frac{1}{-f}=(\frac{2\mu }{3}-1)\frac{2}{R}.......\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$

$-(\frac{2}{3}-1)\frac{2}{R}=(\frac{2\mu }{3}-1)\frac{2}{R}$

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

$\Rightarrow \mu =2$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.