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  • A short linear object of length L lies on the axis of a spherical mirror of focal length f at a distance b from the mirror. The size of the image is:Option: 1</s

A short linear object of length L lies on the axis of a spherical mirror of focal length f at a distance b from the mirror. The size of the image is:

Option: 1

\mathrm{-\frac{f^{2}}{(b-f)^{2}} L}


Option: 2

\mathrm{\frac{f^{2}}{(b-f)^{2}} L}


Option: 3

\mathrm{-\frac{f^{2}}{(b+f)^{2}} L}


Option: 4

\mathrm{-\frac{\mathrm{f}^{2}}{\mathrm{~b}^{2}} \mathrm{~L}}


Answers (1)

best_answer

\mathrm{\frac{1}{v}+\frac{1}{u}=\frac{1}{f}}

By differentiating both sides
\mathrm{\Rightarrow \frac{d v}{v^{2}}=-\frac{d u}{u^{2}}}

du : size of object \mathrm{=\mathrm{L}}
\mathrm{\mathrm{dv}} : size of image
\mathrm{\mathrm{u}}: object distance
\mathrm{\mathrm{v}} : image distance.

\mathrm{ d v=-\frac{v^{2}}{u^{2}} d u . \quad\left(\because \frac{1}{v}=\frac{1}{f}-\frac{1}{u} \quad \Rightarrow \quad v=\frac{f u}{u-f}\right)}
\mathrm{ =-\frac{f^{2} u^{2}}{(u-f)^{2} u^{2}} d u=-\frac{f^{2}}{(b-f)^{2}} L}
Negative sign implies that object is lying between u and \mathrm{ u+du} and the image will lied between v and v-dv.

Posted by

shivangi.shekhar

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