?An object of height 8 cm is kept at a distance of 35 cm in front of a concave mirror. The focal length of the mirror is 25 cm. At what distance from the mirror should a screen be kept so as to get a clear image? What will be the size and nature of the image?

Dear student,
Please find below the answer:

As per the Mirror formula

1/f = 1/v + 1/u

where

f = focal length = -25 cm

v = image distance

u = object distance = -35 cm

so,

-(1/25) = -(1/35) + 1/v

or 1/v = -1/35 + 1/25

or 1/v = 2/175

The image distance is of the distance where the screen should be placed. It will be 87.5 cm. The image will be formed behind the mirror and it will be virtual and erect.

The size of the image will be magnified.

Magnification = -v/u = hi/ho
Where, hi= Height of image
ho= Height of object

hi = (-v/u) X ho

or hi = -(87.5/-35) x 8  

hi = 20 cm

Hence, the image is magnified at 20 cm.

Hope this clears your doubt.
Regards.

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