An object is placed at a distance of 12,cm in | vedclass

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(B) Initially,the object is at $x = -12\,cm$ and the mirror is at $x = 0$. The image is formed at $x = +12\,cm$.When the mirror is moved $4\,cm$ towar

Vedclass · Accepted Answer

$8\,cm$ towards the mirror

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(B) Initially,the object is at $x = -12\,cm$ and the mirror is at $x = 0$. The image is formed at $x = +12\,cm$.When the mirror is moved $4\,cm$ towards the object,the new position of the mirror is $x = -4\,cm$.The distance of the object from the new mirror position is $u = |-12 - (-4)| = 8\,cm$.Since the image is formed at the same distance behind the mirror,the new image position $x_i$ satisfies $|x_i - (-4)| = 8\,cm$,so $x_i = 4\,cm$.The initial position of the image was $12\,cm$ and the final position is $4\,cm$.Therefore,the shift in the position of the image is $12\,cm - 4\,cm = 8\,cm$ towards the mirror.

An object is placed at a distance of $12\,cm$ in front of a plane mirror. $A$ virtual and erect image is formed by the mirror. Now,the mirror is moved by $4\,cm$ towards the stationary object. The distance by which the position of the image would be shifted is:

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