Use the mirror equation to deduce that:
(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
(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

Mirror formula,
`1/f = 1/u + 1/v`
`therefore 1/v = 1/f - 1/u`...... (1)
Now, for concave mirror, `f lt 0` (negative) and as objective is in left to mirror `u lt 0`
`therefore 2 f lt u lt f` .. ... (1)
Now, for concave mirror `f lt 0` (negative) and as object is in left to mirror , `u lt 0`
`therefore 2 f lt u lt f`
`therefore 1/ (2f) gt (1)/(u) gt (1)/(f)`
`therefore (1)/f - (1)/(2f) lt (1)/(f) - (1)/(u) lt (1)/(f) - (1)/(f)`
[ By adding 1/f in each term]
`therefore 1/(2f) lt (1)/(v) lt 0 [ therefore ` from equation (1)]
This represent that `v lt 0,` hence real image is obtained in left of mirror.
`2 f gt 0`
`therefore | 2f | gt | v| [ because 2f` and v are negative ]
Hence, real image is obtained beyond 2f.
For convex lens, ` f gt 0` and object is in left , hence, `u lt 0`
`therefore` From mirror equation ,
`1/v = 1/f - 1/u`
Now,` f lt 0` and `u lt 0` and it represents