Question

The power of a thin convex lens $\left({}_a{n}_g=1.5\right)$ is $5.0D$. When it is placed in a liquid of refractive index ${}_a{n}_l$, then it behaves as a concave lens of focal length $100cm$. 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, $-5=\frac{0.5n}{1.5-n}$
or $-7.5+5n=0.5n$
or $-7.5=-4.5n$
or $n=\frac{75}{45}=\frac{5}{3}$