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Q 9.15 (a) 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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The equation we have for a mirror is:

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

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

Given condition  f<u<2f and v>2f

                          \frac{1}{2f}<\frac{1}{u}<\frac{1}{f}        and  \frac{1}{v}<\frac{1}{2f}

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

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

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

          {2f}<{v}<0

Here f has to be negative in order to satisfy the equation and hence we conclude that our mirror is a concave Mirror. It also satisfies that -v>-2f(image lies beyond 2f)

Posted by

Pankaj Sanodiya

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