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An object `4 cm` in size is placed at a distance of `25.0 cm` from a concave mirror of focal length 15.0 cm. Find the position, nature and height of the image.

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Here, height of object, `h_(1) = +4 cm`
object distance, `u = -25.0 cm`
focal length of concave mirror, `f = -15.0 cm`
Note that plus/minus signs are as per New Cartesian Sign Conventions.
(1) Position of Image. Using the given values in the mirror formula:
`1/v+1/u=1/f` ,
`1/v=1/f-1/u` or `1/v=1/(-15) - 1/(-25) = (-5+3)/75`
or `v = 75/(-2) = -37.5 cm`
i.e., image distance, `v =-37.5 cm.`
Thus, position of image is 37.5 cm in front of the mirror, because v is negative.
(ii) Nature of Image. Since the image is formed in front of the mirror, it must be real and inverted.
Height of Image. Using the expression for linear magnification,
`m = h_(2)/h_(1) = -v/u`
and putting `h_(1) = + 4 cm`, `u = -25.0 cm` and `v = -37.5 cm`, we get
`h_(2)/4 = -((-37.5)/(-25.0)) = (-3)/2` or `h_(2) = -6 cm`
Therefore, height of image is 6 cm. The minus sign shows that the image is below the principal axis. Therefor, it is inverted and real.

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