Question

An object is gradually moving away from the focal point of a concave mirror along the axis of the mirror. The graphical representation of the magnitude of linear magnification \((m)\) versus distance of the object from the mirror \((x)\) is correctly given by :
(Graphs are drawn schematically and are not to scale)

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

For a concave mirror, \(|m| = \frac{f}{x - f}\) for \(x \gt f\). It tends to infinity near \(x = f\), equals 1 at \(x = 2f\) and decreases towards 0 for large \(x\).

Answer 1

The Correct Option is C

Mirror formula: \(\frac{1}{v} + \frac{1}{u} = \frac1f\) with \(u = -x\), \(f \lt 0\) (sign convention). The magnification magnitude is \(|m| = \left|\frac{v}{u}\right| = \frac{f}{x - f}\) in terms of the focal length magnitude \(f\), for \(x \gt f\).
As \(x \to f^+\), \(|m| \to \infty\). At \(x = 2f\), \(|m| = \frac{f}{f} = 1\). For \(x \gg f\), \(|m| \to 0\), decreasing monotonically.
The graph must therefore start very high just to the right of \(x = f\), pass through \(m = 1\) at \(x = 2f\) and keep decreasing. Option (3) shows the curve starting above at \(x \gt f\) and passing through \(m = 1\) at \(x = 2f\), so it is the best match.
Answer: (3) [graph through \((2f, 1)\), decreasing from very large values near \(x = f\)]