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A short straight object of length $l$ lies along the central axis of a spherical concave mirror, at a distance $X$ from the mirror. The focal length of the mirror is $F$. If the length of the image in the mirror is $l^{\prime}$, then ratio $\left(\frac{l^{\prime}}{l}\right)$ is (assume, $l< < $ and $l< < F$ )
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Verified Answer
The correct answer is:
$\left(\frac{F}{F-X}\right)^2$
The given situation is shown in the following figure.
$\begin{aligned}
& A B \rightarrow \text { object, } A^{\prime} B^{\prime} \rightarrow \text { image } \\
& \qquad A B=l, A^{\prime} B^{\prime}=l^{\prime}, u=-X,
\end{aligned}$
Hence, by mirror formula,
$\begin{aligned}
\frac{1}{-F} & =\frac{1}{v}+\frac{1}{(-X)} \quad\left[\because \frac{1}{f}=\frac{1}{v}+\frac{1}{u}\right] \\
\Rightarrow \quad \frac{1}{v} & =\frac{1}{x}-\frac{1}{F} \Rightarrow \frac{1}{v}=\frac{F-X}{F X}
\end{aligned}$
Longitudinal magnification for small object is given as
$\begin{aligned}
M & =\frac{l^{\prime}}{l}=-\frac{v^2}{u^2}=-\frac{\left(\frac{F X}{F-X}\right)^2}{(-X)^2} \\
& =-\frac{f^2 X^2}{(F-X)^2 X^2}=-\left(\frac{F}{F-X}\right)^2
\end{aligned}$
Magnitude of longitudinal magnification,
$\frac{l^{\prime}}{l}=\left(\frac{F}{F-X}\right)^2$
$\begin{aligned}
& A B \rightarrow \text { object, } A^{\prime} B^{\prime} \rightarrow \text { image } \\
& \qquad A B=l, A^{\prime} B^{\prime}=l^{\prime}, u=-X,
\end{aligned}$
Hence, by mirror formula,
$\begin{aligned}
\frac{1}{-F} & =\frac{1}{v}+\frac{1}{(-X)} \quad\left[\because \frac{1}{f}=\frac{1}{v}+\frac{1}{u}\right] \\
\Rightarrow \quad \frac{1}{v} & =\frac{1}{x}-\frac{1}{F} \Rightarrow \frac{1}{v}=\frac{F-X}{F X}
\end{aligned}$
Longitudinal magnification for small object is given as
$\begin{aligned}
M & =\frac{l^{\prime}}{l}=-\frac{v^2}{u^2}=-\frac{\left(\frac{F X}{F-X}\right)^2}{(-X)^2} \\
& =-\frac{f^2 X^2}{(F-X)^2 X^2}=-\left(\frac{F}{F-X}\right)^2
\end{aligned}$
Magnitude of longitudinal magnification,
$\frac{l^{\prime}}{l}=\left(\frac{F}{F-X}\right)^2$
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