Question

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.

Answer 1

The focal lengt `f=-15//2cm = -7.5cm`
i) The object distance `u=-10cm`. Then
Eq. `(1)/(upsilon)+(1)/(u)=(1)/(f)" gives"`
`(1)/(v)+(1)/(-10)=(-1)/(-7.5)`
`or upsilon =(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=-5cm`. Then
from Eq. `(1)/(upsilon)+(1)/(u)=(1)/(f)`
`(1)/(v)+(1)/(-5)=(1)/(-7.5)`
`or upsilon =(5xx7.5)/((7.5-5))=15cm`
This image is formed at 15 cm behind the mirror. It is a virtual image.
Magnification `m=-(upsilon)/(u)=-(15)/((-5))=3`
The image is magnified, vertual and erect.