Question

The focal length of a plano convex lens is '$f$' and its refractive index is $1.5$. It is kept over a plane glass plate with its curved surface touching the glass plate. The gap between the lens and the glass plate is filled by a liquid. As a result, the effective focal length of the combination becomes $2f$. Then the refractive index of the liquid is

• A. $1.25$
  
• B. $1.33$
  
• C. $1.5$
  
• D. $2$
  

Answer 1

The correct option is A $1.25$

From lens maker's formula:-

$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$

For, plano-convex lens:-

$\frac{1}{f}=(1.5-1)(\frac{1}{-\infty }-\frac{1}{-R})=\frac{1}{2R}$

$⟹f=2R$

Now, we can treat the fluid below the plano-convex lens as a plano -concave lens of refractive index $\mu$ and focal length $f^{{}^{\prime }}$

So, $\frac{1}{f^{{}^{\prime }}}=(\mu -1)(\frac{1}{-R}-\frac{1}{\infty })$

$⟹\frac{1}{f^{{}^{\prime }}}=\frac{1-\mu }{R}$

Now, for combination:-

$\frac{1}{F}=\frac{1}{f}+\frac{1}{f^{{}^{\prime }}}$

$\frac{1}{2f}=\frac{1}{f}+\frac{1-\mu }{R}$

$⟹\frac{1-\mu }{R}=\frac{-1}{2f}=\frac{-1}{4R}$

$⟹1-\mu =-0.25$

$⟹\mu =1.25$

Answer-(A)