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  • A short object of length L is placed along the principal axis of a concave mirror away from focus. The object distance is u. If the mirror has a focal length f, what will be the length of the image? You may take L<< | v-f |

A short object of length L is placed along the principal axis of a concave mirror away from focus. The object distance is u. If the mirror has a focal length f, what will be the length of the image? You may take L<<\left | v-f \right |

 

Answers (1)

Thin mirror formula :

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

u= object distance and v= image distance

Since, the object distance is ,u. Let us consider the two ends of the object be at distance u_{1} and u_{2}  respectively, so that du=\left | u_{1}-u_{2} \right |=L. hence size of the image can be written as dv.

By differentiating both sides

\left ( -\frac{du}{u^{2}} \right )+\left ( -\frac{dv}{v^{2}} \right )=0\Rightarrow \frac{dv}{v^{2}}=-\frac{du}{u^{2}}

or

dv=-\left ( \frac{v^{2}}{u^{2}} \right )du \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ...(i)

As \frac{1}{v}=\frac{1}{f}-\frac{1}{u}\Rightarrow v=\frac{fu}{u-f}

Or \frac{v}{u}=\frac{f}{u-f} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ...(ii)

From (i) and (ii

) dv=-\left ( \frac{f}{u-f} \right )^{2}du

But du=L, hence the length of the image is

\frac{f^{2}}{(u-f)^{2}}L

This is the required expression for the length of the image.

Posted by

infoexpert23

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