An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharply focused image can be obtained? Find the size and the nature of the image.

Asked by Vishal kumar | 2 years ago |  431

##### Solution :-

Object distance (u) = - 27 cm

Object height (h) = 7 cm

Focal length (f) = - 18 cm

According to the mirror formula,

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

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

$$=-\frac{1}{18}+\frac{1}{27}=-\frac{1}{54}$$

v = -54 cm

The screen should be placed at a distance of 54 cm in front of the given mirror

$$Magnification,m=-\frac{Image\,Distance}{Object\,Distance}=\frac{-54}{27}=-2$$

The negative value of magnification indicates that the image formed is real

$$Magnification,m=\frac{Height\,of\,the\,image}{Height\,of\,the\,object}=\frac{h^1}{h}$$

$$h^1=m\times h=7\times -2=-14cm$$

The negative value of image height indicates that the image formed is inverted

Answered by Shivani Kumari | 2 years ago

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