An object,4.0 , cm in size,is placed 25.0 , c | vedclass

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(A) Given:Object size,$h = +4.0 \, cm$Object distance,$u = -25.0 \, cm$Focal length,$f = -15.0 \, cm$Using the mirror formula:$\frac{1}{f} = \frac{1}{

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$-37.5 \, cm$,Real,Inverted,$6.0 \, cm$

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(A) Given:Object size,$h = +4.0 \, cm$Object distance,$u = -25.0 \, cm$Focal length,$f = -15.0 \, cm$Using the mirror formula:$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-15} - \frac{1}{-25} = \frac{-5 + 3}{75} = \frac{-2}{75}$$v = -37.5 \, cm$The negative sign indicates that the image is formed in front of the mirror,so a screen must be placed at $37.5 \, cm$ from the mirror on the same side as the object. The image is real.Magnification,$m = \frac{h'}{h} = -\frac{v}{u}$$h' = -\frac{v \cdot h}{u} = -\frac{(-37.5) \cdot (4.0)}{-25.0} = -6.0 \, cm$The image size is $6.0 \, cm$. The negative sign indicates that the image is inverted.

An object,$4.0 \, cm$ in size,is placed $25.0 \, cm$ in front of a concave mirror of focal length $15.0 \, cm$. 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.

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