Question

A liquid of refractive index $1.6$ is contained in the cavity of a glass specimen of refractive index $1.5$ as shown in figure. If each of the curved surfaces has a radius of curvature of $0.20\text{m}$, the arrangement behaves as a diverging lens. The magnitude of focal length of this lens $\left(\text{in cm}\right)$ is

Answer 1

We know that, if lenses are placed in contact, the focal length of the equivalent 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 equi- concave lens is given by
$$f=-\frac{R}{2({\mu }_1-1)}$$
Focal length of an equi - convex lens is given by
$$f=\frac{R}{2({\mu }_2-1)}$$
From the given figure, Using $\left(1\right)$ we can write that,

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

$\Rightarrow f=\frac{R}{2{\mu }_2-4{\mu }_1+2}$

Substituting the given data we get,

$f=\frac{0.2}{(2\times 1.6)-(4\times 1.5)+2}$

$\Rightarrow f=-\frac{0.2}{0.8}=-0.25\text{m}=-25\text{cm}$
Thus, magnitude of focal length is $25\text{cm}$