A short straight object of height 100, cm lie | vedclass

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(D) $1$. The magnification $m$ is given by the ratio of image height to object height: $m = \frac{h_i}{h_o} = \frac{25\, cm}{100\, cm} = +0.25$.$2$. S

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Image is virtual,opposite side of convex mirror.

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(D) $1$. The magnification $m$ is given by the ratio of image height to object height: $m = \frac{h_i}{h_o} = \frac{25\, cm}{100\, cm} = +0.25$.$2$. Since the magnification is positive $(m > 0)$,the image is erect (same orientation as the object) and virtual.$3$. For a spherical mirror,a virtual and erect image is formed by both concave and convex mirrors. However,a concave mirror produces a magnified virtual image $(|m| > 1)$,whereas a convex mirror produces a diminished virtual image $(|m| $4$. Since the image height $(25\, cm)$ is less than the object height $(100\, cm)$,the magnification is less than $1$. Therefore,the mirror must be a convex mirror.$5$. $A$ virtual image formed by a mirror is always located on the opposite side of the mirror relative to the object.

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