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Q 9.15 (c)    Use the mirror equation to deduce that:

                    the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.

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In a convex mirror focal length is positive conventionally.

so we have mirror equation

 \frac{1}{f}=\frac{1}{u}+\frac{1}{v}

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

here since f is positive and u is negative (conventionally) so we have,

 \frac{1}{v}>\frac{1}{f} that is '

v<f 

which means the image will always lie between pole and focus.

Now,

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

magnification (m)=-\frac{v}{u} = \frac{f}{f-u}

here since u is always negative conventionally, it can be seen that magnification of the image will be always less than 1 and hence we conclude that image will always be diminished.

 

 

Posted by

Pankaj Sanodiya

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