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

Here, Object distance, u = - 27 cm (To the left side of mirror)
Image length, v = ? (To be calculated)
Focal length, f = -18 cm (It is a concave mirror)
`"Height of object, "h_(21) = 7.0 cm `
`"And, Height of image, "h_(2)=?` (To be calculated)
`"Now, For a mirror : 1/v+1/u=1/f`
`"So, "1/v+1/(-27)=1/(-18)`
`1/v-1/27=-1/18`
`1/v=-1/18+1/27`
`1/v=(-3+2)/54`
`1/v=-1/54`
Image distance, v = - 54 cm
Since the image distance is minus 54 cm, therefore, the screen should be placed at a distance of 54 cm in front of the concave mirror (on its left side). The nature of image obtained on the screen is real and inverted.
`"Also, For a mirror, "h_(2)/h_(1)=-v/u`
`"So, "h_(2)/(7.0)=-(-54)/(-27)`
`h_(2)/(7.0)= -2`
`h_(2) = -2xx7.2`
`"Height of image ,"h_(2) = -14.0 cm`
Thus, the size of image is 14.0 cm.