Question

A planoconvex lens when silvered on the plane side behaves like a concave mirror of focal length $60cm$. However, when silvered on the convex side it behaves like a concave mirror of focal length $20cm$. Then the refractive index of the lens is

• A. 3.0
• B. 1.5
• C. 1.0
• D. 2.0

Answer 1

Answer: (B)

$\frac{1}{F}=\frac{1}{f_l}+\frac{1}{f_l}+\frac{1}{f_m}$
$\frac{1}{F}=\frac{2}{f_l}+\frac{1}{f_m}$
or $\frac{1}{F}=2(\mu -1)\left(\frac{1}{R}\right)+\frac{1}{\infty }$
or $F=\frac{R}{2\mu (-1)}$
Now, $-60=\frac{-R}{2(\mu -1)}$
or $120(\mu -1)=R$...(1)
Again, $\frac{1}{F}=\frac{1}{f_l}+\frac{1}{f_l}+\frac{1}{f_m}$
or $\frac{1}{F}=\frac{2}{f_l}+\frac{1}{R/2}$

or $\frac{1}{F}=2(\mu -1)\left(\frac{1}{R}\right)+\frac{2}{R}$
or $\frac{1}{F}=\frac{2}{F}(\mu -1+1)$
or $F=\frac{R}{2\mu }$
Now, $-20=\frac{-R}{2\mu }$ or $40\mu =R\dots$(2)
Dividing (1) by (2), we get
$\frac{120(\mu -1)}{40\mu }=\frac{R}{R}=1$
or $120(\mu -1)=40\mu$
or $120\mu -40\mu =120$
or $80\mu =120$
or $\mu =\frac{120}{80}=\frac{3}{2}=1.5$