Question

A planoconvex 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 $20\, cm$. 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 $40 \mu=R \ldots$(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 $\mu=\frac{120}{80}=\frac{3}{2}=1.5$