Question:
A point object in air is in front of the curved surface of a plano-convex lens. The radius of curvature of the curved surface is $30 \mathrm{~cm}$ and the refractive index of the lens material is $1.5$, then the focal length of the lens (in $\mathrm{cm}$ ) is
Solution:
(60) Given : $\mu=1.5 ; R_{\text {curved }}=30 \mathrm{~cm}$
Using, Lens-maker formula
$\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$
For plano-convex lens
$R_{1} \rightarrow \infty$ then $R_{2}=-R$
$\therefore f=\frac{R}{\mu-1}=\frac{30}{1.5-1}=60 \mathrm{~cm}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.