Assume that a certain spherical mirror has a focal length of -10.0 cm. Locate and describe the
image for object distances of (a) 25.0 cm (b) 10.0 cm (c) 5.0 cm.

Explanation

(a) `(1)/(v)+(1)/(u)=(1)/(f)`
`u=-25 cm`
and `f=-10 cm`
`implies(1)/(v)+(1)/(-25)=(1)/(-10)`
`v=-16.7 cm`
`m=(-v)/(u)=-((-16.7)/(-25))=-0.667`
`v` is - ve, means image is formed on the same side of the mirror as the object, image is real. The magnification is - ve, implies the image is inverted. The magnification is less than `1`, hence image is diminished.

(b) `u=-20cm, f=-10cm implies(1)/(v)+(1)/(-20)=(1)/(-10)v=-20 cm`
`m=(-v)/(u)=-((-20)/(-20))=-1`.
The image is real and formed at the same position as the object. When object is at the centre of curvature of a concave mirror, `(u=R=2f=2(-10)=-20 cm)` the image is also formed there itself. `m=-1implies` image is of same size and inverted.
(c ) `u=-15cm` and `f=-10cm`
`(1)/(v)+(1)/(-15)=(1)/(-10)impliesv=-30cm`
Image is real, `m=(-v)/(u)=-((-30)/(-15))=-2`
Image is inverted and enlarged.
(d) `u=-10cm` and `f=-10cm`
`(1)/(v)+(1)/(-10)=(1)/(-10)`
`v=oo`
When the object is at the focus, image is formed at infinity. This means, the reflected rays travel parallel to each other.
`m=(-v)/(u)=-((oo)/(-10))=oo`
(e ) `u=-5cm` and `f=-10cm`
`(1)/(v)+(1)/(-5)=(1)/(-10)impliesv=+10cm`
Image is virtual. It is formed inside the mirror.
`m=(-v)/(u)=-((10)/(-5))=+2`
The image is erect and enlarged.