A short straight object of length l lies alon | vedclass

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(C) The given situation is shown in the figure.Let $AB$ be the object and $A^{\prime}B^{\prime}$ be the image.Here, $AB = l$, $A^{\prime}B^{\prime} =

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$\left(\frac{F}{F-X}\right)^2$

Vedclass · Answer

(C) The given situation is shown in the figure.Let $AB$ be the object and $A^{\prime}B^{\prime}$ be the image.Here, $AB = l$, $A^{\prime}B^{\prime} = l^{\prime}$, and the object distance $u = -X$.Using the mirror formula, $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$, where $f = -F$ for a concave mirror:$\frac{1}{-F} = \frac{1}{v} + \frac{1}{-X}$$\frac{1}{v} = \frac{1}{X} - \frac{1}{F} = \frac{F-X}{FX}$$v = \frac{FX}{F-X}$For a small object placed along the principal axis, the longitudinal magnification $M$ is given by:$M = \frac{l^{\prime}}{l} = -\frac{dv}{du} = -\frac{d}{du} \left( \frac{fu}{u-f} \right) = -\frac{f^2}{(u-f)^2}$Substituting $u = -X$ and $f = -F$:$M = -\frac{(-F)^2}{(-X - (-F))^2} = -\frac{F^2}{(F-X)^2}$The magnitude of the longitudinal magnification is:$\left| \frac{l^{\prime}}{l} \right| = \left( \frac{F}{F-X} \right)^2$

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