Use the mirror equation to deduce that:
an object placed between f and 2f of a concave mirror produces a real image beyond 2f.

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Solution

The mirror formula is

\frac{1}{u} + \frac{1}{v} = \frac{1}{f}

∴

\frac{1}{v} = \frac{1}{f} - \frac{1}{u}

Now for a concave mirror, f < 0 and for an object on the left u < 0.

∴

2 f < u < f

\Rightarrow

\frac{1}{2 f} > \frac{1}{u} > \frac{1}{f}

\Rightarrow

- \frac{1}{2 f} < - \frac{1}{u} < - \frac{1}{f}

\Rightarrow

\frac{1}{f} - \frac{1}{2 f} < \frac{1}{f} - \frac{1}{u} < \frac{1}{f} - \frac{1}{f}

\Rightarrow

\frac{1}{2 f} < \frac{1}{v} < 0

This implies that v < 0 so that real image is formed on left. Also, the above inequality implies that

2 f > v

|2 f| > |v|

[∵ 2 f and v are - ve]

i.e., real image is formed beyond 2f.

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