An 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

Question

An 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.

Accepted Answer

Given:
Object size ($h$) = $+4$ cm
Object distance ($u$) = $-25$ cm (by sign convention)
Focal length ($f$) = $-15$ cm (by sign convention)
Image distance ($v$) = ?
Image size ($h'$) = ? 
Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ 
Substitute 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 
The screen should be placed at a distance of 37.5 cm in front of the concave mirror. 
Now, calculating the magnification ($m$): $m = \frac{h'}{h} = -\frac{v}{u}$ $m = -\frac{-37.5}{-25} = -1.5$ 
Size of the image ($h'$): $h' = m 	imes h = -1.5 	imes 4 = -6$ cm 
Nature of the image: Real and inverted, and its size is 6 cm (enlarged).

Light – Reflection and Refraction

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