Question

The power of a thin convex lens $({}_a{n}_g=1.5)$ 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. $\frac{5}{3}$
  
• B. $\frac{4}{3}$
  
• C. $\sqrt{3}$
• D. $\frac{5}{4}$
  

Answer 1

The correct option is A $\frac{5}{3}$


By using lens maker's formula,
$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$
and we know that power of lens P= 5 D
$\Rightarrow 5=(1.5-1)\frac{2}{R}$......(1)

If a lens of refractive index ${\mu }_g$ is immersed in a liquid of refractive index ${\mu }_l$, then its focal length in liquid:
$\frac{1}{f_l}=({}_l{\mu }_g-1)(\frac{1}{R_1}-\frac{1}{R_2})$
$\Rightarrow -1=(\frac{1.5}{n}-1)\frac{2}{R}$......(2)
Dividing (1) and (2)
$-5=\frac{0.5n}{1.5-n}$
$\Rightarrow -7.5+5n=0.5n\Rightarrow -7.5=-4.5n$
$\Rightarrow n=\frac{75}{45}=\frac{5}{3}$