An object is placed (i)`10 cm` (ii) `5 cm` in front of a convex mirror of radius of curvature `15 cm`. Find the position, nature and magnification of the image in each case.

Explanation

Here, `R = -15 cm`,
`f = ( R)/(2)= -(15)/(2) cm`.
(i) `u = -10 cm, v = ?`
From `(1)/(v)+(1)/(u)=(1)/(f)`
`(1)/(v)=(1)/(f)-(1)/(u)=(-2)/(15) + (1)/(10) = (-4 + 3)/(30) = (-1)/(30)`
`v = -30 cm`.
`:.` This image is at `30 cm` rom the mirror on the same side as object. Also,
magnification, `m = - (v)/(u) = -((-30))/((-10))= -3`
`:.` Time image is real, inverted and magnified `3` times.
(ii) `u = -5 cm, v = ?`
From `(1)/(v)+(1)/(u)=(1)/(f)`
`(1)/(v)=(1)/(f)-(1)/(u)=(-2)/(15) + (1)/(5) = (-4 + 6)/(30) = (2)/(30)`
`v = 15 cm`
`:.` The image is at `15 cm` behind the mirror.
Also, `m = -(v)/(u) = -((15))/(-(5)) = 3`
`:.` The image is virtual, erect and magnified three times.