F= -15cm
m= -3 or+3
case 1
m= -v/u
-3=-v/u
v= 3u----[1]
1/f=1/v+1/u
1/-15=1/3u+1/u ----from eqn. [1]
1/-15=4/3u
3u=-60
u=-60/3cm
CASE 2
3=-v/u
v=-3u----[2]
1/f=1/v+1/u
1/-15=-1/3u+1/u
-1/15=2/3u
-3u=30
u=-10

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Problem: An object is kept in front of a concave mirror of focal length 15cm the image formed is 3 times size of object calculate the position, distance of object from the image?

Solution:

Given:
- Focal length of concave mirror, f = -15 cm (negative sign indicates concave mirror)
- Magnification, m = 3

To Find:
- Position of object, u
- Position of image, v
- Distance between object and image, d

Formula Used:
- Magnification, m = -v/u
- Lens Formula, 1/f = 1/v - 1/u
- Distance between object and image, d = u + v

Calculation:
- Given magnification, m = 3
- Using magnification formula, m = -v/u, we get v/u = -3
- Let's assume that the distance of the object from the concave mirror is u
- Using lens formula, 1/f = 1/v - 1/u, we get 1/-15 = 1/v - 1/u
- Simplifying the above equation, we get v - u = -15/3 = -5
- We know that v/u = -3, so we can write v = -3u
- Substituting v = -3u in v - u = -5, we get -3u - u = -5
- Solving the above equation, we get u = 5/2 cm
- Now, v = -3u = -3 x 5/2 = -15/2 cm
- Distance between object and image, d = u + v = 5/2 - 15/2 = -5 cm (negative sign indicates that the image is formed on the same side as the object)

Answer:
- Position of object, u = 5/2 cm
- Position of image, v = -15/2 cm
- Distance between object and image, d = -5 cm

Therefore, the object is placed at a distance of 5/2 cm in front of the concave mirror and the image is formed at a distance of 15/2 cm behind the mirror. The distance between the object and the image is 5 cm.