Question

If the size of a virtual image formed by a concave mirror of focal length $15\text{cm}$ is twice the size of the object, then size of the object is

• A. $15\text{cm}$
• B. $10\text{cm}$
• C. $7.5\text{cm}$
• D. $5\text{cm}$

Answer 1

The correct option is C

$7.5\text{cm}$

Step 1: Given data and assumptions:

1. Magnification of the image is, $m=2$
2. The focal length of the concave mirror is, $f=-15\text{cm}$
3. Here, the negative sign is due to the sign convention.
4. Assume, the distance of the object from the pole is, $u$
5. According to the sign convention, this distance is negative.
6. The distance of the image from the pole is, $v$
7. Since the image is virtual, this distance is positive.

Step 2: Calculation:

The magnification can be expressed as:

$m=\frac{v}{u}$

Therefore, $v=m\times u=2u$

$u,v\text{and}f$ for a mirror are related as follows:

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

Putting the values:

$$\begin{gathered} \frac{1}{2u}+\frac{1}{-u}=\frac{1}{-f} \\ \frac{1}{f}=\frac{1}{2u} \\ u=\frac{f}{2} \\ u=\frac{15}{2}\text{cm} \\ \text{=7.5 cm} \end{gathered}$$

Therefore, the correct option is (c).