Question

a) 5.0 cm tall object is placed perpendicular to the principal axis of
aconcave mirror of focal length 10 cm. The distance of the object from
the mirror is 15 cm. What is the image distance?
b) Find the nature, position and size of the image formed.
c) Represent the situation with the help of a ray diagram.

Answer 1

distance of image is 30
and size of image is10cm invert and real image formed

Answer 2

Given :
height of object=ho=5cm
focal length=f= -10cm[After sign conventions]
Object distance=u= - 15cm
Image distance =v=?

Formula to be used : 
1/f=1/u+1/v
1/v=1/f-1/u
=- 1/10 +1/15
=-3+2/30
= -1/30
v=-30cm
So when object is placed at a distance of 15cm [between F and C] the image is formed at a distance of 30cm beyond C

Magnification :
m=-v/u
m=30/-15=-2
m=-2
Nature of Image :
-->Real
-->Inverted
-->Magnified

Size of image : 
m=hi/ho
hi=mxho
=-2x5=-10cm

Please refer the attachment for the ray diagram