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 $$= 0.5\,cm$$
B.
Real, inverted, height $$= 4\,cm$$
C.
Real, inverted, height $$= 1\,cm$$
D.
Virtual, upright, height $$= 1\,cm$$
Answer :
Real, inverted, height $$= 4\,cm$$
Solution :
Given, $$R = 20\,cm$$
$${h_0} = 2\,cm\,\,{\text{and}}\,\,u = - 30\,cm$$
In general we have assumed $$\mu = 1.5$$
$$\eqalign{
& {\text{As}}\,\,\frac{1}{f} = \left( {1.5 - 1} \right) \times \frac{2}{{20}} = \frac{1}{{20}} \cr
& {\text{So,}}\,\,f = 20\,cm \cr
& {\text{Now,}}\,{\text{as}}\,\,\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \cr
& \frac{1}{{20}} = \frac{1}{v} + \frac{1}{{30}} \cr
& \frac{1}{v} = \frac{1}{{20}} - \frac{1}{{30}} = \frac{{10}}{{600}} \cr
& \therefore v = 60\,cm \cr
& {\text{As}}\,\,m = \frac{v}{u} = \frac{{60}}{{ - 30}} = - 2 \cr
& \Rightarrow m = \frac{{{h_i}}}{{{h_o}}} = - 2 \cr
& \therefore {h_i} = - 2 \times 2 = - 4\,cm \cr} $$
Here, image is real, inverted, magnified and height of image is $$4\,cm.$$