A point object is placed at a distance $10 c m$ and its real image is formed at a distance of $20 c m$ from concave mirror. If the object is moved by $0.1 c m$ towards the mirror, the image will shift by about

0.4 c m

away from the mirror

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0.4 c m

towards the mirror

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0.8 c m

away from the mirror

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0.8 c m

towards the mirror

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Solution

The correct option is A $0.4 c m$ away from the mirror
$W e k n o w \frac{1}{f} = \frac{1}{u} + \frac{1}{v}; w h e r e f = f o c a l l e n g t h = ? u = i n i t i a l d i s t a n c e = 20 c m v = f i n a l d i s t a n c e = 10 c m \frac{1}{f} = \frac{1}{- 20} + \frac{1}{(- 10)} f = \frac{20}{3} c m N o w o b j e c t m o v e d t o w a r d s t h e m i r r o r = 0.1 c m, s o t h e n e w v a l u e o f u = 9.9 c m A g a i n \frac{1}{f} = \frac{1}{u_{1}} + \frac{1}{v_{1}} \frac{1}{f} - \frac{1}{u_{1}} = \frac{1}{v_{1}} \frac{1}{v_{1}} = \frac{3}{20} - \frac{1}{(- 9.9)} v_{1} = 20.4 c m T h e c h a n g e i n f i n a l d i s t a n c e = 20.4 c m - 20 c m = 0.4 c m v = \frac{9 \times 1600}{3600} = 4 c m / s$

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A point object is placed at a distance 10cm and its real image is formed at a distance of 20cm from concave mirror. If the object is moved by 0.1cm towards the mirror, the image will shift by about

A point object is placed at a distance $10 c m$ and its real image is formed at a distance of $20 c m$ from concave mirror. If the object is moved by $0.1 c m$ towards the mirror, the image will shift by about