To solve this problem, we need to apply the lens formula and the concept of magnification to determine how fast the image is moving as the point source approaches the lens. Let's break it down step by step.

Understanding the Lens Formula

The lens formula is given by:

1/f = 1/v - 1/u

Where:

• f is the focal length of the lens (10 cm in this case).
• v is the image distance from the lens.
• u is the object distance from the lens (which will be negative for real objects).

Setting Up the Problem

Initially, the object (the point source of light) is 15 cm away from the lens. Since we consider distances measured from the lens, we have:

u = -15 cm (the negative sign indicates that the object is on the same side as the incoming light).

Calculating the Image Distance

Using the lens formula:

1/f = 1/v - 1/u

Substituting the known values:

1/10 = 1/v - 1/(-15)

This simplifies to:

1/10 = 1/v + 1/15

To solve for v, we need a common denominator, which is 30:

3/30 = 1/v + 2/30

Rearranging gives:

1/v = 3/30 - 2/30 = 1/30

Thus, we find:

v = 30 cm

Finding the Rate of Change of Image Distance

Now, we need to find how fast the image distance is changing as the object approaches the lens. We know the object is moving towards the lens at a speed of 2 cm/s. We can relate the rates of change of the object distance (u) and the image distance (v) using the lens formula.

Taking the derivative of the lens formula with respect to time (t), we have:

0 = -1/v² (dv/dt) + 1/u² (du/dt)

Here, du/dt is the rate at which the object distance is changing (which is -2 cm/s, since u is decreasing), and dv/dt is what we want to find.

Substituting Values

We already found that:

u = -15 cm and v = 30 cm

Now substituting these values into the derivative equation:

0 = -1/(30)² (dv/dt) + 1/(-15)² (-2)

This simplifies to:

0 = -1/900 (dv/dt) + 1/225 (2)

Multiplying through by 900 to eliminate the fractions gives:

0 = -dv/dt + 4

Thus, we find:

dv/dt = 4 cm/s

Direction of Image Movement

Since the image distance is increasing (as indicated by a positive rate), the image is moving away from the lens. Therefore, the correct answer is:

4 cm/s away from the lens.

Final Answer

The image is moving at a rate of 4 cm/s away from the lens, which corresponds to option C.