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Q.

A converging lens is to projected image of a lamp 4 times the size of the lamp on a wall at a distance of 10 m from the lamp. The focal length of the lens is

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a

1.6m

b

2.67m

c

4.4m

d

-1.6m

answer is A.

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Detailed Solution

$\begin{array}{l} m = \frac{v}{u} \\ - 4 = \frac{v}{u} or v = - 4 u \\ Now, v - u = 10 or - 5 u = 10 \\ or u = - 2 m \\ v = 8 cm \\ Again, \frac{1}{f} = \frac{1}{v} - \frac{1}{u} or \\ \frac{1}{f} = \frac{1}{8} - \frac{1}{- 2} or \frac{1}{f} = \frac{1}{8} + \frac{1}{2} = \frac{1 + 4}{8} = \frac{5}{8} or \\ f = \frac{8}{5} m = 1 .6 m \end{array}$

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