Optics Ques 24
- 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) $(b \frac{u-f^{1 / 2}}{f})$
(b) $(b \frac{f}{u-f})$
(c) $(b \frac{u-f}{f})$
(d) $(b \frac{f^{2}}{u-f})$
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Answer:
Correct Answer: 24.(d)
Solution:
Formula:
- From the mirror formula
$ \begin{aligned} \frac{1}{v}+\frac{1}{u} & =\frac{1}{f} \quad(f=\text { constant }) \quad …….(i) \\ -v^{-2} d v-u^{-2} d u & =0 \\ |d v| & =\frac{v^{2}}{u^{2}}|d u| \quad …….(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})^2 $
Substituting these values in Eq. (ii), we get
Size of image $=b (\frac{f}{u-f})^2$
$\therefore \quad $ Correct option is (d).
BETA
