Question

A biconvex lens has a radius of curvature of magnitude $20\, cm$. Which one of the following options describe best the image formed of an object of height $2 \,cm$ placed $30\, cm$ from the lens?

• A. Virtual, upright, height = 1 cm
• B. Virtual, upright, height = 0.5 cm
• C. Real, inverted, height = 4 cm
• D. Real, inverted, height = 1 cm

Answer 1

Answer: (C)

Given,
Radius of curvature, $R=20 cm$
Height of object, $h _0=2$
Object distance, $u =30 cm$
We have, $\frac{1}{ f }=(\mu-1)\left(\frac{1}{ R _1}-\frac{1}{ R _2}\right)$
$ \begin{array}{l} =\left(\frac{3}{2}-1\right)\left[\frac{1}{20}-\left(-\frac{1}{20}\right)\right] \\ \Rightarrow \frac{1}{ f }=\left(\frac{3}{2}-1\right) \times \frac{2}{20} \end{array} $
$ \therefore f =20 cm $
$ \frac{1}{ f }=\frac{1}{ v }-\frac{1}{ u } $
$ \Rightarrow \frac{1}{20}=\frac{1}{ v }+\frac{1}{30} $
$ \frac{1}{ v }=\frac{1}{20}-\frac{1}{30} $
$ =\frac{10}{600} $
$ v =60 cm $
$ m =\frac{ h _{ i }}{ h _0}=\frac{ v }{ u } $
$ \Rightarrow h _{ i }=\frac{ v }{ u } \times h _0 $
$ =\frac{60}{30} \times 2=-4 cm $
so, image is inverted.