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# Question and Answer

## QuestionPhysicsClass 12

An object is placed at $$10 \ \text{cm}$$  in front of a concave mirror of radius of curvature $$15 \ \text{cm}$$ . Find the position, nature, and magnification of the image.

Given data,
Radius of curvature $$R=-15\text{ cm}$$
Distance of object $$u=-10\text{ cm}$$
Focal length $$f=\frac R 2$$
$$=-\frac{15} 2$$
$$=-7{\cdot}5\text{ cm}$$
By mirror formulae,
$$\frac 1 v+\frac 1 u=\frac 1 f$$ , by putting values
$$\Rightarrow \frac 1 v+\left(\frac{-1}{10}\right)=\left(\frac{-1}{7{\cdot}5}\right)$$
$$\Rightarrow v=-30\text{ cm}$$
$$-\mathit{ve}$$  sign indicates that image is real and inverted and it’s position is $$30\text{ cm}$$  from the mirror.
Magnification,
$$m=\frac{-v} u$$
$$=\frac{-30}{-10}$$
$$=3$$
Thus the image is $$3$$ times magnified.
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