A concave mirror has a radius of curvature of 24 cm. How far is
an object from the mirror if an image is formed that is:
(a) virtual and 3.0 times the size of the object,
(b) real and 3.0 times the size of the object and
(c) real and `1//3` the size of the object?.

Explanation

Given,R=-24 cm
Hence, `f=R/2=-12 cm`
(i) image is virtual and three times larger. Hence, u is negative and v is positive. Simultaneousely, `abs(v)=3abs(u)`. So let,
u=-x
then, v=+3x
Substituting in the mirror formula `1/v+1/u=1/f`, we have
`1/(3x)-1/x=1/(-12)`
`therefore` x=8 cm
Therefore, object distance is 8 cm.
(ii) Image is real and three times larger. Hence, u and v both are negative and `abs(v)=3(u)`. So let,
u=-x
then, v=-3x
Substituting in mirror formula, we have
`1/(-3x)-1/x=-1/12or x=16 cm`
`therefore` Object distance should be 16 cm.
(iii) Image is real and `1/3`rd the size of object. Hence both u and v are negative and `abs(v)=abs(u)/3`. So let,
u=-x
then,
`v=-x/3`
Substituting in the mirror formula, we have
`-3/x-1/x=-1/12`
x=48 cm
`therefore` Object distance should be 48 cm.