For a concave mirror,if a real image is forme | vedclass

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(A) The mirror formula is given by $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.Rearranging this to express $\frac{1}{v}$ in terms of $\frac{1}{u}$,we ge

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(A) The mirror formula is given by $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.Rearranging this to express $\frac{1}{v}$ in terms of $\frac{1}{u}$,we get $\frac{1}{v} = -\frac{1}{u} + \frac{1}{f}$.This equation is of the linear form $y = mx + c$,where $y = \frac{1}{v}$,$x = \frac{1}{u}$,slope $m = -1$,and intercept $c = \frac{1}{f}$.For a concave mirror,the focal length $f$ is negative,so $c = \frac{1}{f}$ is negative. However,for a real image,$u$ and $v$ are both negative. As $u$ varies from $f$ to $\infty$,$v$ varies from $\infty$ to $f$. The graph is a straight line with a slope of $-1$ and a negative intercept on both axes,which corresponds to the form shown in option $A$.

For a concave mirror,if a real image is formed,the graph between $\frac{1}{u}$ and $\frac{1}{v}$ is of the form:

Similar Questions

$A$ convex mirror and a concave mirror of radius $10\; cm$ each are placed $15\; cm$ apart facing each other. An object is placed midway between them. If the reflection first takes place in the concave mirror and then in the convex mirror,the position of the final image is:

For a concave mirror, if a virtual image is formed, the graph between $m$ and $u$ is of the form:

In the figure shown,consider the first reflection at the plane mirror and the second at the convex mirror. $AB$ is the object.

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