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 $$4\,cm.$$

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Optics and Wave >> Ray Optics

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A. $${\sin ^{ - 1}}\left( {\tan r} \right)$$
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