Optics 1 Question 9
9. A short linear object of length $b$ lies along the axis of a concave mirror of focal length $f$ at a distance $u$ from the pole of the mirror. The size of the image is approximately equal to
(a) $b\left(\frac{u-f}{f}\right)^{1 / 2}$
(b) $b\left(\frac{f}{u-f}\right)^{1 / 2}$
(c) $b\left(\frac{u-f}{f}\right)$
(d) $b\left(\frac{f}{u-f}\right)^2$
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Answer:
Correct Answer: 9. (d)
Solution:
- From the mirror formula
$$ \begin{aligned} \frac{1}{v}+\frac{1}{u} & =\frac{1}{f} \quad(f=\text { constant })\cdots(i) \\ -v^{-2} d v-u^{-2} d u & =0 \\ or \quad |d v| & =\frac{v^{2}}{u^{2}}|d u| \cdots(ii) \end{aligned} $$
Here, $|d v|=$ size of image
$|d u|=$ size of object (short) lying along the axis $=b$
Further, from Eq. (i), we can find
$$ \frac{v^{2}}{u^{2}}=\frac{f}{u-f} $$
Substituting these values in Eq. (ii), we get
Size of image = $b\left(\frac{f}{u-f}\right)^2$
$\therefore$ Correct option is (d).
