CBSE 2025, Set 3

Question

Using the mirror equation and the formula of magnification, deduce that "the virtual image produced by a convex mirror is always diminished in size and is located between the pole and the focus."

Edvaya · Accepted Answer

The correct answer is Subjective. Let the object be placed at a distance $u$ in front of the convex mirror. By sign convention, $u$ is negative. Let $u = -nf$, where $f$ is the focal length (which is positive for a convex mirror) and $n$ is any positive number.

From the mirror formula:
$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$
$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$$

Substituting $u = -nf$:
$$\frac{1}{v} = \frac{1}{f} - \frac{1}{-nf} = \frac{1}{f} + \frac{1}{nf}$$
$$\frac{1}{v} = \frac{n+1}{nf}$$
$$v = \frac{nf}{n+1}$$

Since $n$ is positive, $v$ is positive, meaning the image is virtual and formed behind the mirror. 
Also, since $\frac{n}{n+1} 0$, $m$ is positive (erect image) and $m < 1$. Therefore, the image is always diminished in size.

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Physics > Optics > Ray Optics And Optical Instruments

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