Question

Diameter of aperture of a plano-convex lens is $6cm$ and thickness at the centre is $3mm$. If the speed of light in the material of the lens is $2\times 10^{8}m{s}^{-1}$, the focal length of the lens is

• A. $15cm$
• B. $20cm$
• C. $30cm$
• D. $10cm$

Answer 1

Answer: (C)

Here, $r=\frac{6}{2}=3cm$, $t=3mm=0.3cm$
If $R_2$ is radius of curvature of convex surface , then $2R_{2}t=r^2$
$\therefore R_2=\frac{r^2}{2t}=\frac{3\times 3}{2\times 0.3}=15cm,R_1=\infty$
and $\mu =\frac{c}{\upsilon }=\frac{3\times 10^8}{2\times 10^8}=\frac{3}{2}$
As $\frac{1}{f}=\left(\mu -1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$\therefore \frac{1}{f}=\left(\frac{3}{2}-1\right)\left(\frac{1}{\infty }-\frac{1}{15}\right)=\frac{1}{30}$
$\Rightarrow f=30cm$