# An arrow 2.5 cm high is placed at a distance of 25 cm from a diverging mirror of focal length 20 cm. Find the nature, position and size of the image f

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An arrow 2.5 cm high is placed at a distance of 25 cm from a diverging mirror of focal length 20 cm. Find the nature, position and size of the image formed.

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Here, object size, h_(1) = +2.5 cm, object distance, u = -25 cm
focal length of diverging mirror, f = +20 cm
image distance, v = ?, image size, h_(2) = ?
Using mirror formula, 1/u + 1/v = 1/f or 1/v = 1/f - 1/u
and putting, u = -25 cm and f = +20 cm, we get
1/v = 1/20 -1/(-25) = (5+4)/100 = 9/100 or v = 100/9 = 11.1 cm
Since image is at the back of the mirror at 11.1 cm from the mirror, it must be virtual and erect. If h_(2) is the size of the image, then as
m = h_(2)/h_(1) = (-v)/u, h_(2)/(2.5) = (-11.1)/(-25) or h_(2) = (11.1)/25 xx 2.5 = 1.11 cm
This is the size of the image.