The image of an illuminated square is obtained on a screen with the help of a converging lens

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:

Smart Achievers · Accepted Answer

(4)

$If side of object square = l$

$and side of image square = l'$

$From question, \frac{l'^{2}}{l^{2}} = 9 or \frac{l'}{l} = 3$

$i.e. magnification m = 3,$

$u = - 40 cm$

$v = 3 \times 40 = 120 cm$

$f = ?$

$From formula, \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{120} - \frac{1}{- 40} = \frac{1}{f}$

$Or, \frac{1}{f} = \frac{1}{120} + \frac{1}{40} = \frac{1 + 3}{120}$

$∴$ $f = 30 cm$

Solution

Q.
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:

1 36 cm

2 27 cm

3 60 cm

4 30 cm

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