Given:
Distance of object from pole, u = -15 cm
Radius of curvature, R = -20 cm (negative sign indicates concave mirror)
Amplitude of oscillation, A = 2 mm

To find:
Amplitude of image, A'

Formula used:
Magnification, m = -v/u
where v is the distance of image from the pole

Derivation:
Let O be the object, C be the center of curvature and F be the focus of the concave mirror. The distance of the object from the pole is u = -15 cm. The radius of curvature is R = -20 cm. The distance of the focus from the pole is f = R/2 = -10 cm.

The amplitude of oscillation of the object is A = 2 mm. This means that the object moves back and forth along the principal axis between two extreme positions, each at a distance of A/2 = 1 mm from the central position.

When the object is at the central position, its image coincides with the object itself. When the object moves to one of the extreme positions, its image moves to a new position. Let I be the image of the object when it is at one of the extreme positions.

Analysis:
1. Position of the object when it is at one of the extreme positions:
When the object is at one of the extreme positions, its distance from the pole is u' = u + A/2 = -14.8 cm. This is because the amplitude of oscillation is much smaller than the distance of the object from the pole, so we can assume that the distance of the object from the mirror remains constant during the oscillation.

2. Position of the image when the object is at one of the extreme positions:
To find the position of the image, we need to first find the distance of the image from the pole, v. This can be done using the formula for magnification:
m = -v/u
where m is the magnification of the mirror.

The magnification of a concave mirror is negative, which means that the image is inverted. The magnification can be found using the mirror formula:
1/f = 1/u + 1/v
where f is the focal length of the mirror.

Substituting the values we get,
1/-10 = 1/-15 + 1/v
v = -30 cm

Substituting this value of v in the magnification formula, we get
m = -v/u
m = -(-30)/(-15) = 2

This means that the image is twice the size of the object and is inverted. The distance of the image from the pole is v = -30 cm.

When the object is at one of the extreme positions, its image is also at a new position. Let A' be the amplitude of oscillation of the image.

3. Position of the image when the object is at the central position:
When the object is at the central position, its image coincides with the object itself. Therefore, its distance from the pole is also u = -15 cm.

Using the same formula for magnification, we can find the distance of the image from the pole when the object is at the central position:
m = -v/u
v = -mu
v = -2*(-15) = 30 cm

This means that when the object is at the central position, its