Thank you for reporting, we will resolve it shortly
Q.
A plano-convex lens has refractive index $1.5$ and radius of curvature $50\, cm$. What is focal length of lens?
Jharkhand CECEJharkhand CECE 2003
Solution:
If $R_{1}$ and $R_{2}$ are the radii of curvature of first and second refracting
surfaces of a thin lens with optical centre $C$ of focal length $f$ and refractive index $\mu$ then according to lens Makers formula
$\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$
where $\mu$ is refractive index of material of lens. with respect to surrounding medium.
For plano-convex lens, $R_{1}=50\, cm , R=\infty$ (for plane surface)
$\frac{1}{f}=(1.5-1)\left(\frac{1}{50}-\frac{1}{\infty}\right) $
or $\frac{1}{f}=0.5 \times \frac{1}{50}$ or
$f=\frac{50}{0.5}=100\, cm$