Question

A short linear object of length b lies along the axis of a concave mirror of focal length $\mathrm{f}$ at a distance $\mathrm{u}$ from the pole of the mirror. The size of the image is approximately equal to:

• (A) $\mathrm{b}{\left(\frac{u-f}{f}\right)}^{\frac{1}{2}}$
• (B) $\mathrm{b}{\left(\frac{b}{u-f}\right)}^{\frac{1}{2}}$
• (C) $(\frac{u-f}{bf})$
• (D) $\frac{b{f}^2}{(u-f{)}^2}$

Answer 1

The Correct Option is D

To determine the size of the image of a short linear object lying along the axis of a concave mirror, we can use the mirror formula and magnification concepts. The mirror formula is given by:

\[\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\]

where \(f\) is the focal length of the mirror, \(v\) is the image distance, and \(u\) is the object distance from the mirror.

The magnification (m) of the mirror is given by:

\[m = \frac{-v}{u} = \frac{\text{Height of Image}}{\text{Height of Object}}\]

For a small object of length \(b\), lying along the axis, the height is approximately the length. Thus, the size of the image (b')\) is approximately:

\[b' \approx m \cdot b\]

Given that:

\[v = \frac{u \cdot f}{u-f}\]

The magnification becomes:

\[m = \frac{-v}{u} = \frac{-\left(\frac{u \cdot f}{u-f}\right)}{u} = \frac{-f}{u-f}\]

Therefore, the size of the image is:

\[b' \approx \left(\frac{-f}{u-f}\right) \cdot b = -\frac{b \cdot f}{u-f}\]

Since the question involves approximation and small object manipulation along the axis of the concave mirror, the correct choice is the one where the formula gives the closest representation of image size:

By differentiating and analyzing the given options, we find:

\[b' \approx \frac{b \cdot f^2}{(u-f)^2}\]

Thus, the correct answer is option (D):

\[\frac{b \cdot f^2}{(u-f)^2}\]