Geometrical OpticsHard

Question

Distance between an object and three times magnified real image is 40 cm . The focal length of the mirror used is $\_\_\_\_$ cm .

Options

A.$- 15/2$
B.-10
C.-20
D.-15

Solution

$m = - 3 = \frac{v}{u}$

$${v = - 3u }{|v| - |u| = 40 }{u = 20\text{ }cm }{v = 60\text{ }cm }{\therefore\frac{1}{v} + \frac{1}{u} = \frac{1}{f} }{\frac{1}{- 60} + \frac{1}{- 20} = \frac{1}{f} }{f = - 15\text{ }cm}$$

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