Assume that an object is kept at a distance of 20 cm in front of a con

1.	Assume that an object is kept at a distance of 20 cm in front of a concave mirror. If its focal length is 30 cm, then a) what is the image distance? b) what the magnification of mirror in this case?
Answer» Object distance = u = 20 cm a) Focal length =f =-30 cm (\(\because\) concave mirror ) Images distance = v = ? Mirror formula \(\frac{1}{f}= \frac{1}{v}\,+\,\frac{1}{u}\Rightarrow\,-\frac{1}{30} = \frac{1}{v}\,-\,\frac{1}{20}\) \(\frac{1}{v} =\frac{1}{20}\,-\,\frac {1}{30}\,\Rightarrow\, \frac{1}{v}=\frac{3\,-\,2}{60}= \frac{1}{60}\Rightarrow v = 60\,cm. \) Image distance = 60 cm. So image is virtual formed in the mirror. b) Magnification = m = \(-\frac{Image \,distance}{Object\,distance} = \frac{-60}{-20}=3\,cm\) Here '+' represents image is erect and it is 3 times enlarged .

Assume that an object is kept at a distance of 20 cm in front of a concave mirror. If its focal length is 30 cm, then a) what is the image distance? b) what the magnification of mirror in this case?

Answer»

Object distance = u = 20 cm

a) Focal length =f =-30 cm (\(\because\) concave mirror )

Images distance = v = ?

Mirror formula \(\frac{1}{f}= \frac{1}{v}\,+\,\frac{1}{u}\Rightarrow\,-\frac{1}{30} = \frac{1}{v}\,-\,\frac{1}{20}\)

\(\frac{1}{v} =\frac{1}{20}\,-\,\frac {1}{30}\,\Rightarrow\, \frac{1}{v}=\frac{3\,-\,2}{60}= \frac{1}{60}\Rightarrow v = 60\,cm. \)

Image distance = 60 cm.

So image is virtual formed in the mirror.

b) Magnification = m = \(-\frac{Image \,distance}{Object\,distance} = \frac{-60}{-20}=3\,cm\)

Here '+' represents image is erect and it is 3 times enlarged .