Q. A short linear object of length $b$ lies on the axis of a concave mirror of focal length $f$ at a distance $u$ from the pole. The length of the image will be.

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A

$\left(\frac{f}{u - f}\right)b$

B

$\left(\frac{u - f}{f}\right)b$

C

$\left(\frac{f}{u - f}\right)^{2}b$

D

$\left(\frac{u - f}{f}\right)^{2}b$

Solution:

The concave mirror formula is
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ .... (i)
Since, for a given concave mirror, focal length $f$ is fixed, we partially differentiate Eq. (i) to get
$-\frac{\delta v}{v^{2}}=-\frac{\delta u}{u^{2}}=0$ or $\deltav=-\left(\frac{v}{u}\right)^{2} \, \deltau$ ...... (ii)
Multiplying Equation. (i) by $u$ , we get
$\frac{u}{v}+1=\frac{u}{f}$ or $\frac{v}{u}=\frac{f}{u - f}$ ..... (iii)
Using Equation. (iii) in Equation. (ii), we get
$\deltav=-\left(\frac{f}{u - f}\right)^{2}\deltau$
Given $\deltau=b.$ Therefore
$\deltav=-\left(\frac{f}{u - f}\right)^{2}b$
The negative sign shown that image is longitudinally inverted. The magnitude of the size of the image is
$\left|\delta v\right|=b\left(\frac{f}{u - f}\right)^{2}$

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