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) For concave mirror, the focal length (f) is negative.
`flt0`
when the object is placed on the left side of the mirror, the object distance (u) is negative.
ult0
For imge distance v, we can write the lens formula as:
`1/v-1/u=1/f`
`1/v=1/f=1/u` ................(1)
The object lies between f and 2.
`therefore 2fltultf` (`therefore ` u and f are negative)
`1/(2f)gt1/ugt1/f`
`-1/2fgt1/ugt1/f`
`1/f-1/(2f)lt1/f-1/ult0`.............(2)
Uisng eq. 1) we get, `1/(2f)lt1/vlt0`
`1/v` is negative, i.e., v is negative.
`1/(2)lt1/v`
`-vgt-2f`
therefore, the image lies beyond 2f.
b) For a convex mirror, the focal length (f) is positive.
`fgt0`
When the object is placed on the left of the mirror, the object distance (u) is negative
`mult0`
For image distance, v we have the mirror formula.
`1/v+1/u=1/f`
`1/v=1/f-1/u`
But we have `ult0`
`therefore 1/v gt 1/f`
`vltf`
Hence,the image formed is diminshed and is located between the focus (f) and the pole.
d) For a concave mirror, the focal length (f) is negative .
ult0
it is placed between the focus (f) and the pole.
`therefore fgtugt0`
`1/flt1/ult0`
`1/f-1/ult0`
For image distance v, we have the mirror formula.
`1/v+1/u=1/f`
`1/v=1/f-1/u`
`therefore 1/v lt0`
`vgt0`
the image is formed on the right side of the mirror. Hence, it is a virtual image. For `ult0` and `vgt0`, we can write. `1/ugt1/v`
`vgtu`
Magnification, `m=v/ugt1`
Hence, the formed image is enlarged.