Prove that when an object is placed in front of a concave lens, the image formed is (i) virtual (ii) erect (iii) diminished (iv) and located between pole and focus.

Explanation

Let an object be placed at a distance x from the centre of a concave lens of focal length f.
u = -x
Focal length = -f
By lens formula,
`(1)/(v) - (1)/(u) = (1)/(f)`
`(1)/(v) = (1)/(f) + (1)/(u)`
`rArr (1)/(c) = -(1)/(f) - (1)/(x)`

`rArr (1)/(v) = -((1)/(f)+(1)/(x))`
`rArr v = -y`
`rArr` v is negative
(i) `v lt 0`, image is formed in front of the lens (i.e., virtual)
(ii) As `m = (v)/(u), m = (-y)/(-x) = (y)/(x)`
`rArr = m gt 0`
(iii) `(1)/(y) = (1)/(f) + (1)/(x)`
`rArr (1)/(y) gt (1)/(x)` or `y lt x`
`rArr m lt 1` or image is smaller in size (i.e., diminished)
(iv) `(1)/(y) = (1)/(f) + (1)/(x)`
`rArr (1)/(y) gt (1)/(f)` or `y lt f`
`:.` image is formed between C and F.