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

                    an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.

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The focal length f of concave mirror is always negative.

Also conventionally object distance u is always negative.

So we have mirror equation:

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

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

Now in this equation whenever u<f , \frac{1}{v}  will always be positive which means v is always positive which means it lies on the right side of the mirror which means image is always virtual.

Now, 

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

 since the denominator is always less than the numerator, so the magnitude  magnification will always be greater than 1

Hence we conclude that image is always gonna be enlarged.

Hence an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.

Posted by

Pankaj Sanodiya

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