Question
The image of an illuminated square is obtained on a screen with the help of a converging lens. The distance of the square from the lens is $$40\,cm.$$ The area of the image is 9 times that of the square. The focal length of the lens is :
A.
$$36\,cm$$
B.
$$27\,cm$$
C.
$$60\,cm$$
D.
$$30\,cm$$
Answer :
$$30\,cm$$
Solution :
If side of object square $$ = \ell $$
and side of image square $$ = \ell '$$
From question, $$\frac{{\ell '}}{\ell } = 9\,\,{\text{or}}\,\,\frac{{\ell '}}{\ell } = 3$$
i.e, magnification $$m = 3$$
$$u = - 40\,cm$$
$$v = 3 \times 40 = 120\,cm\,f = ?$$
From formula, $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$
$$\eqalign{
& \frac{1}{{120}} - \frac{1}{{ - 40}} = \frac{1}{f}\,\,{\text{or,}}\,\,\frac{1}{f} = \frac{1}{{120}} + \frac{1}{{40}} = \frac{{1 + 3}}{{120}} \cr
& \therefore f = 30\,cm \cr} $$