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Q. An object is placed at $9\, cm$ in front of a concave mirror of radius of curvature $12\, cm$. The following statement is true

KEAMKEAM 2019Ray Optics and Optical Instruments

Solution:

Given, object distance $u=-9\,cm$, radius of curvature, $R=-12\,cm$
As we know mirror formula
$\frac{1}{v}+\frac{1}{u} =\frac{2}{R} \Rightarrow \frac{1}{v}-\frac{1}{9}=\frac{2}{12} $
$\Rightarrow v=-18\,cm$
Now, lateral magnification
$m=-\frac{v}{u}=-\frac{(-18)}{(-9)}=-2$
So, image is magnified, real and inverted.