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 convex 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

(a) Using mirror formula, `(1)/(v)=(1)/(f)-(1)/(u)`
Now, for a concave mirror, `f lt 0` and for an object on the left, `mu lt 0`
`therefore 2f lt mu lt u or (1)/(2f)gt (1)/(u) gt (1)/(f) Rightarrow -(1)/(2f) lt -(1)/(u) lt -(1)/(f) `
`Rightarrow (1)/(f)-(1)/(2f)lt (1)/(f)-(1)/(u) lt (1)/(f)-(1)/(f) therefore (1)/(2f)lt (1)/(v)lt 0`
`Rightarrow v lt 0` so that image is formed on left.
Also, the above inequality implies `2f gt v`
`|2f| lt |v|" "....[therefore "2f and v are -ve"]`
i.e., the real image is formed beyond 2f.
(b) For a convex mirror, `f gt 0` and for an object on left `u lt 0`.
From the mirror formula, `(1)/(v)=(1)/(f)-(1)/(u)`
`Rightarrow (1)/(v) gt 0 or v gt 0`
This shows that whatever be the value of u, a convex mirror forms a virtual image on the right.
(c ) From mirror formula, `(1)/(v)=(1)/(f)-(1)/(u)`
For a concave mirror, `f lt 0` and for an object between the pole and focus of a concave mirror `f lt u lt 0`.
`therefore (1)/(f) gt (1)/(u)" "or (1)/(f)-(1)/(u) gt0" "or (1)/(v)gt0`
i.e., a virtual image is formed on the right
`"Also", (1)/(v) lt (1)/(|u|)" "or v gt |u|" "therefore |m|=(v)/(|u|)gt 1`
i.e. image is enlarged.