Question

The power of a thin convex lens $\left(_{a} n_{g}=1.5\right)$ is $5.0\, D$. When it is placed in a liquid of refractive index ${ }_{a} n_{l}$, then it behaves as a concave lens of focal length $100\, cm$. The refractive index of the liquid ${ }_{a} n_{l}$ will be

• A. 5/3
• B. 4/3
• C. $\sqrt{3}$
• D. 5/4

Answer 1

Answer: (A)

Here, $5=(1.5-1)\left(\frac{2}{R}\right)$
If a lens (made of glass) of refractive index $\mu_{g}$ is immersed in a liquid of refractive index $\mu_{l}$, then its focal length in liquid $f_{l}$ is given by
$\frac{1}{f_{l}}=\left(\mu_{g}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]$
$-1=\left(\frac{1.5}{n}-1\right)\left(\frac{2}{R}\right)$
Dividing, $-5=\frac{0.5 n}{1.5-n}$
or $-7.5+5 n =0.5 n$
or $-7.5=-4.5 n$
or $n=\frac{75}{45}=\frac{5}{3}$