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Q 9.1  A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm.  At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?

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Given, size of the candle, h = 2.5 cm

            Object distance, u = 27 cm

            The radius of curvature of the concave mirror, R = -36 cm

           focal length of a concave mirror = R/2 = -18 cm

           let image distance = v 

now, as we know 

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

\frac{1}{-27}+\frac{1}{v}=\frac{1}{-18}

\frac{1}{v}=\frac{1}{-18} + \frac{1}{27}

v= -54cm

now, let the height of image be  h'

 magnification of the image is given by 

m=\frac{h'}{h}=-\frac{v}{u}

from here

{h'}=-\frac{-54}{-27}*2.5=-5cm

Hence the size of the image will be -5cm. negative sign implies that the image is inverted and real

if the candle is moved closer to the mirror, we have to move the screen away from the mirror in order to obtain the image on the screen. if the image distance is less than the focal length image cannot be obtained on the screen and image will be virtual.

Posted by

Pankaj Sanodiya

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