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  • A thin rod of length  is placed along the optical axis of a concave mirror of focal

A thin rod of length \mathrm{f / 3}  is placed along the optical axis of a concave mirror of focal length \mathrm{f } such that its image which is real and elongated just touches the rod. Calculate the magnification.

Option: 1

1


Option: 2

1.5


Option: 3

2


Option: 4

2.5


Answers (1)

best_answer

Let \ell be the length of the image.
\text { Then, } \mathrm{m}=\frac{\ell}{\mathrm{f} / 3} \Rightarrow \ell=\frac{\mathrm{mf}}{3}



Also image of one end coincides with the object,

\Rightarrow \mathrm{u}^{\prime}=2 \mathrm{f}
u^{\prime}=\mathrm{u}+\frac{f}{3} \Rightarrow \mathrm{u}=2 \mathrm{f}-\frac{f}{3}=\frac{5 \mathrm{f}}{3}
\mathrm{v}=-\left(u+\frac{f}{3}+\frac{\mathrm{mf}}{3}\right). Putting in mirror formula,
\mathrm{\frac{1}{u+f / 3+\mathrm{mf} / 3}+\frac{1}{u}=\frac{1}{f}}
\mathrm{\Rightarrow \frac{3}{5 \mathrm{f}+f+\mathrm{mf}}+\frac{3}{5 \mathrm{f}}=\frac{1}{f} \Rightarrow \frac{1}{\mathrm{~m}+6}=\frac{2}{15} }
\mathrm{\Rightarrow \mathrm{m}=\frac{3}{2}=1.5 }
 

 

Posted by

Rakesh

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