An object of height 4 cm is kept at a distance of 30 cm from the pole of a diverging mirror. If the

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An object of height 4 cm is kept at a distance of 30 cm from the pole of a diverging mirror. If the focal length of the mirror is 10 cm , the height of the image formed is\

Professor PC Thomas Classes · Answer

(c): Given, height of object $(h)=+4 \mathrm{~cm}$ Object distance $(u)=-30 \mathrm{~cm}$ (object placed left side of the mirror)\ Focal length, $f=+10 \mathrm{~cm}$\ Mirror formula, $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$ or $v=\frac{u f}{u-f}$\ $v=\frac{(-30) 	imes(10)}{(-30)-(10)} \quad$ or $\quad v=\frac{30}{4}=7.5 \mathrm{~cm}$\ Now, magnification $(m)=\frac{-v}{u}=\frac{h^{\prime}}{h} ; m=-\frac{\left(\frac{15}{2}ight)}{-30}=\frac{h^{\prime}}{4}$\ or $\quad h^{\prime}=1 \mathrm{~cm}$\ Hence, height of the image formed is 1 cm .\