Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

The graph that shows the correct variation of $\frac{1}{v}$ with $\frac{1}{u}$ for a concave mirror, where $u$ is the object distance and $v$ is the image distance, is:

[Image of Graph 1]

[Image of Graph 2]

[Image of Graph 3]

[Image of Graph 4]

Step-by-Step Solution

Mirror Formula: The relationship between object distance ( $u$ ), image distance ( $v$ ), and focal length ( $f$ ) for a spherical mirror is given by the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ .
Equation of the Line: Rearranging this formula to plot $\frac{1}{v}$ on the y-axis and $\frac{1}{u}$ on the x-axis gives $\frac{1}{v} = -\frac{1}{u} + \frac{1}{f}$ . This represents the equation of a straight line, $y = mx + c$ , where the slope $m = -1$ and the y-intercept $c = \frac{1}{f}$ .
Sign Convention: For a concave mirror, the focal length $f$ is negative. Therefore, both the y-intercept ( $y = \frac{1}{f}$ ) and the x-intercept ( $x = \frac{1}{f}$ when $y=0$ ) must be on the negative axes.
Conclusion: The correct graph is a straight line with a slope of $-1$ that passes through the negative x-axis and negative y-axis (i.e., a straight line in the second, third, and fourth quadrants intersecting the negative axes).

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