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

Explanation

The focal length `f=-15//2cm=-7.5cm`
(i) The object distance `u=-10cm.` Then Eq. (9.7) gives
`(1)/(v)+(1)/(-10)=(1)/(-7.5)`
`"or "v=(10xx7.5)/(-2.5)=-30cm`
The image is 30 cm from the mirror on the same side as the object.
`" Also, magnification m "=-(v)/(u)=-((-30))/(-10)=-3`
The image is magnified, real and inverted.
(ii) The object distance `u = -5 cm`. Then from Eq. (9.7),
`(1)/(v)+(1)/(-5)=(1)/(-7.5)`
`or v=(5xx7.5)/((7.5-5))=15cm`
This image is formed at 15 cm behind the mirror. It is a virtual image.
Magnification `m=-(v)/(u)=-(15)/((-5))=3`
The image is magnified, virtual and erect.