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  • A boy of height stands in front of a convex mirror. His distance from the mirror is equal t

A boy of height \mathrm{1\ m} stands in front of a convex mirror. His distance from the mirror is equal to its focal length. The height of his image is:

Option: 1

\mathrm{0.25\ m}


Option: 2

\mathrm{0.33\ m}


Option: 3

\mathrm{0.5\ m}


Option: 4

\mathrm{0.67\ m}


Answers (1)

best_answer

Let \mathrm{f} be focal length of the convex mirror. According to new cartesian sign convention.
Object distance, \mathrm{u=-f} , focal length \mathrm{=+f}
According to mirror formula
\mathrm{\begin{array}{ll} \mathrm{\frac{1}{u}+\frac{1}{v}=\frac{1}{f}} & \\\\ \mathrm{\frac{1}{-f}+\frac{1}{v}=\frac{1}{f}} & \mathrm{\text { or } \quad \frac{1}{v}=\frac{1}{f}+\frac{1}{f}=\frac{2}{f} \quad \text { or } \quad v=\frac{f}{2}} \end{array}}
The image is formed at a distance \mathrm{\frac{f}{2}} behind the mirror. It is a virtual image.
\mathrm{\text { Magnification, } M=-\frac{v}{u}=-\frac{(f / 2)}{(-f)}=\frac{1}{2}}
\mathrm{\begin{aligned} & \text { Also, }\mathrm{ M}=\frac{\text { Height of image }\left(\mathrm{h}_1\right)}{\text { Height of } \operatorname{object}\left(\mathrm{h}_{\mathrm{o}}\right)} \\ & \therefore \quad \mathrm{h}_{\mathrm{l}}=\mathrm{Mh}_{\mathrm{O}}=\frac{1}{2}(1 \mathrm{~m})=0.5 \mathrm{~m} \end{aligned}}

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Ritika Jonwal

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