An object of size `7.0 cm` is placed at `27 cm` in front of a concave mirror of focal length `18 cm`. At what distance from the mirror, should a screen be placed, so that a sharp focussed image can be obtained ? Find the size and nature of the image ?

Explanation

Object distance, u = -27 cm
Object height, h = 7 cm
Focal length f = -18 cm
Using the mirror formula
`1/f=1/v+1/u`
`1/v=1/f-1/u=(1)/((-18))-(1)/((-27))`
`1/v=(1)/((-18))+(1)/((27))=((-3+2))/(54)= -(1)/(54)`
v= -54 cm
The screen should be placed at a distance of 54 cm in front of the given mirror.
Magnification, `m= ("Image distance")/("Object distance")=(-54)/( -27)` = -2
Negative value of magnification shows that image is real.
Magnification, `m= -("Height of image")/("Height of Object")=(h.)/(h)`
`implies h.=m xx h=(-2)xx7= -14 cm`
The height of image is negative which shows that image is inverted.