Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. An extended object is placed at point $O, 10 \,cm$ in front of a convex lens $L_1$ and a concave lens $L_2$ is placed $10\, cm$ behind it, as shown in the figure. The radii of curvature of all the curved surfaces in both the lenses are $20 \,cm$. The refractive index of both the lenses is $1.5$. The total magnification of this lens system isPhysics Question Image

JEE AdvancedJEE Advanced 2021

Solution:

$\frac{1}{ f _{1}}=(1.5-1)\left(\frac{1}{20}+\frac{1}{20}\right)=\frac{1}{20}$
$\frac{1}{ f _{2}}=(1.5-1)\left(-\frac{1}{20}-\frac{1}{20}\right)=-\frac{1}{20}$
So, $\frac{1}{ v }-\frac{1}{-10}=\frac{1}{20}$
So, $v =-20\, cm$
and $\frac{1}{v^{\prime}}-\frac{1}{-30}=\frac{1}{-20}$
So, $v^{\prime}=-12 \,cm$
So total magnification $=\left(\frac{-20}{-10}\right)\left(\frac{-12}{-30}\right)=0.8$