Question

A plano-convex lens, when silvered on the plane side, behaves like a concave mirror of focal length $60$ cm. 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)

$1.5$

$\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_1}+\frac{1}{f_m}$
$\frac{1}{F}=\frac{2}{f_1}+\frac{2}{f_m}$
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 $120(\mu -1)=R...\left(i\right)$
Again, $\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_1}+\frac{1}{f_m}$
or $\frac{1}{F}=\frac{2}{f_1}+\frac{1}{R/2}$
or $\frac{1}{F}=2(\mu -1)\left(\frac{1}{R}\right)+\frac{2}{R}$
or $\frac{1}{F}=\frac{2}{R}(\mu -1+1)$
or $F=\frac{R}{2\mu }$
Now, $-20=\frac{-R}{2\mu }$
or $40\mu =R....\left(ii\right)$
Dividing (i) by (ii), 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 $\mu =\frac{120}{80}=\frac{3}{2}=1.5$.