A wire is bent in the shape of a right-angled | vedclass

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(B) The wire is placed between $f/2$ and $f$. For a concave mirror,the mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Let a point $P$ on

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(B) The wire is placed between $f/2$ and $f$. For a concave mirror,the mirror formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Let a point $P$ on the hypotenuse be at a distance $x$ from the focus $f$,so $u = -(f-x)$.Substituting this into the mirror formula: $\frac{1}{v} - \frac{1}{f-x} = -\frac{1}{f}$.$\frac{1}{v} = \frac{1}{f-x} - \frac{1}{f} = \frac{x}{f(f-x)}$.Thus,$v = \frac{f(f-x)}{x}$.The magnification $M = -\frac{v}{u} = -\frac{f(f-x)/x}{-(f-x)} = \frac{f}{x}$.For the vertical leg of the triangle at $u = -f/2$,the image is at $v = -f$ (real,inverted,magnified by $2$).For the point at the focus $(u = -f)$,the image is at infinity $(v \rightarrow \infty)$.As $x$ increases from $0$ to $f/2$,the magnification $M = f/x$ decreases from $\infty$ to $2$. The image of the hypotenuse is a curve that extends to infinity,consistent with option $B$.

$A$ wire is bent in the shape of a right-angled triangle and is placed in front of a concave mirror of focal length $f$,as shown in the figure. Which of the figures shown in the four options qualitatively represent$(s)$ the shape of the image of the bent wire? (These figures are not to scale.)

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