For a concave mirror,the frac{1}{u} versus fr | vedclass

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(D) The mirror equation is given by $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Rearranging this into the form of a straight-line equation $y = mx + c$,

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(D) The mirror equation is given by $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Rearranging this into the form of a straight-line equation $y = mx + c$,we get $\frac{1}{v} = -\frac{1}{u} + \frac{1}{f}$.Comparing this with $y = mx + c$,where $y = \frac{1}{v}$ and $x = \frac{1}{u}$,the slope $m = -1$ and the intercept $c = \frac{1}{f}$.From the graph,the intercept on the $\frac{1}{v}$ axis (when $\frac{1}{u} = 0$) is $-0.10 \ cm^{-1}$.Therefore,$\frac{1}{f} = -0.10 \ cm^{-1}$.Calculating the focal length,$f = \frac{1}{-0.10} = -10 \ cm$.

For a concave mirror,the $\frac{1}{u}$ versus $\frac{1}{v}$ graph obtained by the parallax method is as shown in the figure. The graph is a straight line with a slope of $-1$. The estimated value of the focal length of the mirror is: (in $cm$)

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