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.

Explanation

Object-size, h = + 4.0 cm,
Object-distance, u = – 25.0 cm,
Focal length, f = –15.0 cm,
Image-distance, v = ?
Image-size, h ′ = ?
From Eq. (10.1):
`1/v+1/u=1/f`
or, `1/v=1/f-1/u=1/-15.0-1/-25.0= -1/15.0+1/25.0`
or, `1/v=(-5.0+3.0)/75.0=(-2.0)/75.0" or, "v= -37.5 cm`
The screen should be placed at 37.5 cm in front of the mirror. The image is real.
Also, magnification, `m=(h')/h= -v/u`
or, `h'= -(vh)/u=((-37.5cm)(+4.0cm))/((-25.0cm))`
Height of the image, h′ = – 6.0 cm The image is inverted and enlarged.