Understanding Concave Lens Image Formation

A concave lens is a diverging lens, meaning it spreads out light rays that pass through it. By convention in optics:

• Light travels from left to right.
• The optical center (O) of the lens is the origin.
• Distances to objects on the left (real objects) are negative.
• Distances to real images on the right are positive.
• Distances to virtual images on the left are negative.
• Focal length ($f$) for a concave lens is always negative.

When an object is placed at infinity (the furthest possible distance), the rays entering the lens are parallel. A concave lens diverges these parallel rays such that they appear to originate from its principal focus (F) on the same side as the object. Therefore, the image formed is always virtual, erect, diminished, and located at the principal focus.

Applying the Lens Formula

We can confirm this using the lens formula, which relates the object distance ($u$), image distance ($v$), and focal length ($f$):

$$ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} $$

From the question, we have:

• Object is placed at infinity, so the object distance $u = -\infty$.
• Focal length of the concave lens $f = -15$ cm (negative because it's a concave lens).

Calculation Steps:

1. Substitute the given values into the lens formula: $$ \frac{1}{v} - \frac{1}{-\infty} = \frac{1}{-15} $$
2. Recall that $\frac{1}{-\infty}$ is equal to 0. So the equation becomes: $$ \frac{1}{v} - 0 = \frac{1}{-15} $$
3. Simplify the equation: $$ \frac{1}{v} = \frac{1}{-15} $$
4. Solve for $v$: $$ v = -15 \text{ cm} $$

Interpreting the Result

The calculated image distance is $v = -15$ cm.

• The negative sign (-) indicates that the image is virtual and formed on the same side of the lens as the object.
• The magnitude of the distance is 15 cm.

This means the virtual image is formed 15 cm away from the lens, on the same side as the object placed at infinity. This corresponds to the principal focal point of the concave lens.