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?
Answer & explanation
Correct answer: option 2
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, $h'=2×h$
$=2×3=6cm$