A point object is placed at a distance of 10, | vedclass

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(A) Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.Given $u = -10\,cm$ and $v = -20\,cm$ (real image).$\frac{1}{f} = \frac{1}{-20

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$0.4\,cm$ away from the mirror

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(A) Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.Given $u = -10\,cm$ and $v = -20\,cm$ (real image).$\frac{1}{f} = \frac{1}{-20} + \frac{1}{-10} = \frac{-1-2}{20} = -\frac{3}{20} \Rightarrow f = -\frac{20}{3}\,cm$.When the object moves $0.1\,cm$ towards the mirror,the new object distance is $u' = -(10 - 0.1) = -9.9\,cm$.Using the mirror formula again: $\frac{1}{v'} = \frac{1}{f} - \frac{1}{u'} = -\frac{3}{20} - (\frac{1}{-9.9}) = -\frac{3}{20} + \frac{10}{99}$.$\frac{1}{v'} = \frac{-297 + 200}{1980} = -\frac{97}{1980} \Rightarrow v' \approx -20.41\,cm$.The image shifts from $20\,cm$ to $20.4\,cm$ away from the mirror. Thus,the shift is $0.4\,cm$ away from the mirror.

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

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