Use the mirror equation to show 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 ) An object placed between the pole and focus of a concave mirror produces a virtual and an enlarged image.

Explanation

`(1)/( v) + ( 1)/( u ) = ( 1)/( f)`
for concave mirror, `f lt 0 ` or f = - ve
for convex mirror,` f gt 0 ` or `f = + ve `
For concave mirror `:`
Let ` f = - c `
Also, Let `u = nf = - nc `
` ( 1)/( v ) = ( 1)/( f) - ( 1)/( u ) = - ( 1)/ ( c ) + ( 1)/( nc ) = ( - n +1 )/( nc )`
`:.` `v = ( nc)/((1-n))`
(a) When object is betweeen f and 2f, we have `1 lt n lt 2 `
`:. ` v is -ve ` rArr `real image
( for n = 1 and n = 2 ), magnitude of v becomes `oo` and 2c, respectively
`:.` Real image is formed beyong 2F .
(b) For convex mirror `:`
`f = + d , `Let u = -pd ( p can have any value )
`(1)/(v ) = ( 1)/( d) = ( 1)/( pd) = ((1+p))/( pd)`
`v = ( pd)/( ( p+_1))`
`:.` v is alsways `+ve` and always less than d.
`:.` Convex mirror always produces a virtual image between pole and focus.
(c ) Object between pole and F we have `0 lt n lt 1` v is `+ve rArr` ( virtual image ) and `|v| gt c `
`:.` We get a virtual, and an enlarged image.