**Introduction**
When an object is placed at a distance from a concave mirror, an image is formed due to the reflection of light. The size and position of the image depends on the distance of the object from the mirror, the focal length of the mirror and the nature of the mirror.

**Given information**
- The object is placed at a distance of 10 cm from the mirror.
- The image formed is half the size of the object.

**Solution**
To find the position of the image, we need to use the mirror formula:

1/v + 1/u = 1/f

where,
v = distance of the image from the mirror
u = distance of the object from the mirror
f = focal length of the mirror

We know that the image formed is half the size of the object. This means that the magnification of the image is 1/2.

Magnification = -v/u = 1/2

Since the magnification is negative, the image is inverted.

Substituting the given values in the above formula, we get:

1/v + 1/(-10) = 1/f
1/v - 1/10 = 1/f

Multiplying both sides by 10v, we get:

10 - v = 10f/v

Multiplying both sides by v, we get:

10v - v^2 = 10f

Rearranging the terms, we get:

v^2 - 10v + 10f = 0

We know that the image is real and inverted, which means that v is negative. Substituting this value in the above equation, we get:

v^2 + 10v + 10f = 0

Using the quadratic formula, we get:

v = (-10 ± √(100 - 40f))/2

Since v is negative, we take the negative sign in the above formula. Substituting the value of magnification, we get:

-10 + √(100 - 40f) = -2u

Simplifying the equation, we get:

u + 5 = √(25 - 10f)

Squaring both sides, we get:

u^2 + 10u + 25 = 25 - 10f

Simplifying the equation, we get:

u^2 + 10u + 10f = 0

Using the quadratic formula, we get:

u = (-10 ± √(100 - 40f))/2

Since the object is placed at a distance of 10 cm from the mirror, we take the positive sign in the above formula. Substituting the value of u, we get:

u = -5 + √(25 - 10f)

Squaring both sides, we get:

u^2 = 25 - 10f - 10u

Substituting the value of u^2 in the mirror formula, we get:

1/v + 1/(-10) = 1/f

1/v - 1/10 = 1/[(25 - 10f - 10u)/u]

Simplifying the equation, we get:

v = (-10u)/3

Substituting the value of u, we get:

v = (-50 + 5√(25 - 10f))/3

**Conclusion

Given:

u= -10cm

h= 2h'

To find: image distance (v)

Solution:

m= h' ÷ h

m= h' ÷ 2h'                    { h=2h' }

m = 1/2

Also;

m = -v/u

1/2 = -v/-10

1/2 =v/10

v =5cm