Use the mirror equation to deduct that :
(a) an object between `f` and `2f` of a concave mirror produces a real image beyond `2 f`.
(b) a convax mirror always produces a virtual image independent of the location of the object.
( c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.
(d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.

Explanation

The focal length (f) of a concave mirror is negative `(f lt 0)`.
The object distance (u) is negative for the object placed on the left - hand side of the mirror `(u lt0)`.
From mirror formula
`(1)/(v)=(1)/(f)-(1)/(u)" …(i)"`
As `flt0` and `u lt 0`
`rArr u lt0`, therefore, the image formed is real.
Since `" "f lt u lt 2f`
`(1)/(f) gt (1)/(u) gt (1)/(2f)`
`-(1)/(f) lt (-1)/(u) lt (-1)/(2f)`
`(1)/(f)-(1)/(2f)lt(1)/(f)-(1)/(u) lt 0" ....(ii)"`
From relation (i) and (ii)
`(1)/(2f) lt (1)/(v) lt 0`
`v gt 2f`. Therefore, the image is formed beyond 2f.
(b) The focal length (f) of the convex mirror is positive `(f gt0)`.
From mirror formula,
`(1)/(u)=(1)/(f)-(1)/(u)`
as `f gt0` and `u lt 0`
`rArr" " v gt 0`
Therefore, the image is formed behind the mirror for any location of the object . Thus, the image is virtual.
(c ) From mirror formula
`(1)/(v)=(1)/(f)-(1)/(u)`
as `f gt0 and u lt 0`
`rArr" "(1)/(v) gt (1)/(f)`
`rArr" "v lt f`
Therefore, the image is formed between the focus and pole.
Therefore, image is always diminished.
(d) From mirror formula
`(1)/(v)=(1)/(f)-(1)/(u)`
The object is placed between the pole and focus of the concave mirror.
`f lt 0, u lt 0`
and `f lt u`
`rArr" "(1)/(v)=((1)/(f)-(1)/(u))gt0`
`therefore" "u gt 0`
As v is positive, the image is formed behind the mirror. So, it will be virtual in nature. As `v gt|u|` so, the image will be enlarged.