An object 1 cm high is placed at 10 cm infront of a concave mirror of focal length 15cm. the height of image is

1 cm

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2 cm

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3 cm

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4 cm

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Solution

The correct option is C 3 cm
$- \frac{1}{f} = \frac{- 1}{4} + \frac{1}{v} - \frac{1}{15} = \frac{- 1}{10} + \frac{1}{v} - \frac{1}{15} + \frac{1}{10} = \frac{1}{v}; \frac{- 10 + 15}{150} = \frac{1}{V} \frac{5}{150} = \frac{1}{V} \frac{h_{I}}{h_{0}} = \frac{V}{u} \frac{h_{I}}{h_{0}} = - \frac{30}{10} h_{I} = 3 c m$

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Find the size distance and Nature of image formed of an object size 1cm height a distance 15cm infront of concave mirror of focal length.

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