In the case of a spherical mirror,if the obje | vedclass

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(B) The mirror formula is given by: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Rearranging this equation in the form of a linear equation $y = mx + c$,

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$A$ straight line not passing through the origin

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(B) The mirror formula is given by: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Rearranging this equation in the form of a linear equation $y = mx + c$,we get: $\frac{1}{v} = -\frac{1}{u} + \frac{1}{f}$.Here,$y = \frac{1}{v}$,$x = \frac{1}{u}$,slope $m = -1$,and intercept $c = \frac{1}{f}$.Since the intercept $c = \frac{1}{f}$ is non-zero for a spherical mirror,the graph of $\frac{1}{v}$ versus $\frac{1}{u}$ is a straight line that does not pass through the origin.

Column-$I$	Column-$II$
$1$. $m = -2$	a. Convex mirror
$2$. $m = -1/2$	b. Concave mirror
$3$. $m = +2$	c. Real image
$4$. $m = +1/2$	d. Virtual image

In the case of a spherical mirror,if the object distance $(u)$ and image distance $(v)$ are measured from the pole,then the graph plotted between $(1/u)$ and $(1/v)$ will be:

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What is the nature of the image formed for an object placed on the principal axis of a concave mirror between the pole and the focus?

Match the corresponding entries of Column-$I$ with Column-$II$. [Where $m$ is the magnification produced by the mirror]
Column-$I$ Column-$II$
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$2$. $m = -1/2$ b. Concave mirror
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$4$. $m = +1/2$ d. Virtual image

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