To calculate the image distance ($v$) for a concave mirror, we use the mirror formula:

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

where:

• $f$ is the focal length of the mirror,
• $v$ is the image distance,
• $u$ is the object distance.

We know:

• Focal length $f = -10 \, \text{cm}$ (negative for concave mirrors),
• Object distance $u = -15 \, \text{cm}$ (since it's on the same side as the object).

Now, we plug these values into the mirror equation:

$\frac{1}{-10} = \frac{1}{v} + \frac{1}{-15}$

Solving this equation:

$\frac{1}{v} = \frac{1}{-10} + \frac{1}{15}$

First, let's find a common denominator and calculate the result:

$\frac{1}{v} = \frac{-3}{30} + \frac{2}{30} = \frac{-1}{30}$

Thus,

$v = -30 \, \text{cm}$

So, the image distance is $v = -30 \, \text{cm}$, meaning the image is 30 cm in front of the mirror on the same side as the object.

Would you like more details or have any questions?

Here are 5 related questions to think about:

1. How does the sign convention affect the interpretation of object and image distances?
2. What happens to the image distance if the object is moved closer to the focal point?
3. How does the magnification change as the object distance varies?
4. How can we calculate the magnification using the image and object distances?
5. How would the result change if the mirror were convex?

Tip: For concave mirrors, images can be real or virtual depending on the object's position relative to the focal point.