CBSE · Class 10 · Science · Light – Reflection and RefractionAn object 4 cm in size is placed at 25 cm in front of a concave mirror of focal length 15 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Find the nature and the size of the image.
Given:\nObject size ($h$) = $+4$ cm\nObject distance ($u$) = $-25$ cm (by sign convention)\nFocal length ($f$) = $-15$ cm (by sign convention)\nImage distance ($v$) = ?\nImage size ($h'$) = ? \nUsing the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ \nSubstitute the given values: $\frac{1}{-15} = \frac{1}{v} + \frac{1}{-25}$ $\frac{1}{v} = -\frac{1}{15} + \frac{1}{25}$ $\frac{1}{v} = \frac{-5 + 3}{75}$ $\frac{1}{v} = \frac{-2}{75}$ $v = -\frac{75}{2} = -37.5$ cm \nThe screen should be placed at a distance of 37.5 cm in front of the concave mirror. \nNow, calculating the magnification ($m$): $m = \frac{h'}{h} = -\frac{v}{u}$ $m = -\frac{-37.5}{-25} = -1.5$ \nSize of the image ($h'$): $h' = m \times h = -1.5 \times 4 = -6$ cm \nNature of the image: Real and inverted, and its size is 6 cm (enlarged).