A double convex lens made of glass of refractive index $\frac{3}{2}$ has both radii of curvature of magnitude 20 cm. An object 3 cm high is placed at 10 cm from the lens. What is the nature and size of the image?

6 cm; virtual and erect

The correct answer is Option (2) → 6 cm; virtual and erect

Using lens maker formula,

$\frac{1}{f}=(u-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$=(1.5-1)\left(\frac{1}{20}-\frac{1}{-20}\right)$

$=\frac{1}{20}⇒f=20cm$

Using the lens equation -

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$⇒\frac{1}{20}=\frac{1}{v}-\frac{1}{-20}$

$⇒v=-20cm$

and,

Magnification = $\frac{v}{u}=\frac{-20}{-10}=2$

∴ Image height, $h'=2×h$

$=2×3=6cm$