An object O is placed 30 cm in front of a concave mirror of focal length 10 cm - Leaving Cert Physics - Question d - 2005

From the question, we know that:

• The focal length, (f = -10) cm (negative for a concave mirror).
• The object distance, (u = -30) cm (negative in the case of mirrors as the object is in front of the mirror).

Now substituting the values into the mirror formula:

$\frac{1}{-10} = \frac{1}{-30} + \frac{1}{v}$

Rearranging the equation, we have:

$\frac{1}{v} = \frac{1}{-10} + \frac{1}{30}$

To find (\frac{1}{v}), we need a common denominator:

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

Thus,

$v = -15\ \text{cm}$

The negative sign indicates that the image is formed on the same side as the object, which is typical for a virtual image in a concave mirror.

Therefore, the image is formed 15 cm from the mirror. Since the image distance is negative, it confirms that it is a virtual image. The final answer can be expressed as:

• The distance of the image from the mirror is 15 cm.