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

(a) The mirror equation is `(1)/(v)+(1)/(u)=(1)/(f)` or `(1)/(v)=(1)/(f)-(1)/(u)`
For a concave mirror, `f` is positive i.e., `f lt 0`. As object is on the left, `u` is negative, i.e., `u lt 0`
As object lies between `f` and `2 f` of a concave mirror, `:. 3f lt u lt f`
`(1)/(2f) gt (1)/(u)gt(1)/(f)` or `-(1)/(2f)lt -(1)/(u) lt -(1)/(f)` or `(1)/(f)-(1)/(2 f) lt (1)/(f) - (1)/(u) lt0` or `(1)/(2f) lt (1)/(v) lt 0`
`:. (1)/(v)` is negative or `v` negative. The image is real. Also `vgt2 f` i.e., the image lies beyond`2 f`
(b) For a convex mirror, `f` is positive i.e., `f gt 0`. As object is on the left, `u` is neagtive, i.e., `u lt 0`
As `(1)/(v)=(1)/(f)-(1)/(u)`
`:. (1)/(v)` is positive or `v` is positive i.e., image is at the back of the mirror. Hence image is virtual, whatever be the value of `u`.
( c) For a convex mirror, `f gt 0 and u lt 0`
As `(1)/(v) =(1)/(f)-(1)/(u)`, therefore `((1)/(v)) gt ((1)/(f))` i.e., `v lt f`
`:.` Image is located between the pole and the focus. As `v lt |u|`, the image is diminished.
(d) For a concave mirror, `f lt 0`
As object is placed between the pole and focus `:. f lt u lt 0 :. ((1)/(f) - (1)/(u)) gt 0`
But `((1)/(f) - (1)/(u)) = (1)/(v) gt 0 or v` is positive. Image is on the right. It must be virtual.
Also, `(1)/(v)lt(1)/(|u|)` i.e., `v gt |u|` `:.` Image is enlarged.