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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, what is the size of image?
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Verified Answer
The correct answer is:
$\left(\frac{f}{u-f}\right)^2 b$
Using the relation for the focal length of concave mirror
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$ ...(1)
Differentiating equation (1), we obtain
$0=-\frac{1}{v^2} d v-\frac{1}{u^2} d u$
So, $\quad d v=-\frac{v^2}{u^2} \times b$ $\ldots(2)$
(Here : $d u=b$ )
From equation (1)
$\begin{aligned}
\frac{1}{v} & =\frac{1}{f}-\frac{1}{u}=\frac{u-f}{f u} \\
\frac{u}{v} & =\frac{u-f}{f} \\
\end{aligned}$
$\frac{v}{u}=\frac{f}{u-f}$ $\ldots(3)$
Now, from equations (2) and (3), we get
$d v=-\left(\frac{f}{u-f}\right)^2 b$
Therefore, size or image is $=\left(\frac{f}{u-f}\right)^2 b$
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$ ...(1)
Differentiating equation (1), we obtain
$0=-\frac{1}{v^2} d v-\frac{1}{u^2} d u$
So, $\quad d v=-\frac{v^2}{u^2} \times b$ $\ldots(2)$
(Here : $d u=b$ )
From equation (1)
$\begin{aligned}
\frac{1}{v} & =\frac{1}{f}-\frac{1}{u}=\frac{u-f}{f u} \\
\frac{u}{v} & =\frac{u-f}{f} \\
\end{aligned}$
$\frac{v}{u}=\frac{f}{u-f}$ $\ldots(3)$
Now, from equations (2) and (3), we get
$d v=-\left(\frac{f}{u-f}\right)^2 b$
Therefore, size or image is $=\left(\frac{f}{u-f}\right)^2 b$
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