An object is kept in front of a concave mirror of focal length 15 cm. The image formed is three times the size of the object. Calculate the two possible distances of the object from the mirror.

Explanation

Concave mirror , f=-15 cm , `h_2=3, h_(1), u= ?`
(a) If the image is virtual then `h_(2)=+3h_1` (`:'` virtual image is errect )
As we know, `(h_(2))/(h_(1))=(-v)/u rArr (3h_(1))/(h_(1))=(-v)/u rArr v=-3u` ..........(By using `h_(2)=3h_(1)`)
Here, `1/v+1/u=1/f rArr 1/(-3u) +1/u=1/(-15) rArr (-1)/(3u) +1/u =(-1)/15` ltbr. `=(-1+3)/(3u)=(-1)/15 rArr 2/(3u) =(-1)/15 rArr (3u)/2 =(-15)/1`
`u=(-15xx2)/3=-10 cm ` (object is placed 10 cm front of the mirror)
(b) If the image is real then `h_2=-3h_1` (real image is inverted)
`(h_2)/(h_1)=(-v)/u rArr =(-3h_1)/(h_1)=(-v)/u rArr v=3u, ` Now, `1/v+1/u =1/f rArr =1/(3u)+1/u =1/(-15) rArr (1+3)/(3u) =(-1)/15`
`4/(3u) =(-1)/15 rArr =15xx4/3 rArr u=-20 cm ` (Object is placed 20 cm in fron tof the mirror)