Question

For the lens arrangement shown in the figure, The effective focal length is $-60\text{m}$. The radii of curvature of the curved surfaces of plano-convex lenses are $12\text{cm}$ each and refractive index of the material of the lens is $1.5$. Find the refractive index of the liquid.

• A. $1.42$
• B. $1.53$
• C. $1.33$
• D. $1.60$

Answer 1

The correct option is D $1.60$

We know that, if lenses are placed in contact, the effective focal length of the lens can be given as,
$$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}.......\left(1\right)$$
Focal length of an plano - convex lens is given by
$$f=\frac{R}{{\mu }_1-1}$$
Focal length of an equi- concave lens is given by
$$f=-\frac{R}{2({\mu }_2-1)}$$
Where ${\mu }_2$ is refractive index of liquid.

Given, Effective focal length , $f=-60\text{cm}$

Radius of curvature of plano - convex lens $R=12\text{cm}$

Refractive index of the lens material ${\mu }_1=1.5$

From the figure, using $\left(1\right)$ we can write that,

$\frac{1}{f}=\frac{{\mu }_1-1}{R}-\frac{2({\mu }_2-1)}{R}+\frac{{\mu }_1-1}{R}$

$\Rightarrow \frac{1}{f}=\frac{2({\mu }_1-1)}{R}-\frac{2({\mu }_2-1)}{R}$

$\Rightarrow \frac{1}{f}=\frac{2({\mu }_1-{\mu }_2)}{R}$

$\Rightarrow {\mu }_2={\mu }_1-\frac{R}{2f}$

Substituting the given data we get,

${\mu }_2=1.5-\frac{12}{2\times (-60)}$

$\Rightarrow {\mu }_2=1.5-(-0.1)=1.6$

Hence, option (d) is the correct answer.