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  • In normal adujstment, for a refracting telescope, the distance between objective and eye piece is <img alt="30 \mathrm{~cm}" src="https://entrancecorner.oncodecogs.com/gif.latex?30%20%5Cmathrm%7B%7

In normal adujstment, for a refracting telescope, the distance between objective and eye piece is 30 \mathrm{~cm}. The focal length of the objective, when the angular magnification of the telescope is 2 , will be:

Option: 1

20\: \mathrm{cm}


Option: 2

30\: \mathrm{cm}


Option: 3

10\: \mathrm{cm}


Option: 4

15\: \mathrm{cm}


Answers (1)

best_answer

In normal adjustment final image is at infinity

\mathrm{\mu _{e}=f_{e}}

Distance between objective and eyepiece\mathrm{=f_{o}+f_{e}}

\mathrm{30=f_{o}+f_{e}}\rightarrow (1)

Angular magnification = \mathrm{\frac{f_{o}}{f_{e}}}

                           \mathrm{2=\frac{f_{o}}{f_{e}}}

\mathrm{f_{e}=\frac{f_{o}}{f_{e}}}\rightarrow (2)

From eq (1) & (2)

\mathrm{30=\frac{3f_{o}}{2}}

\mathrm{f_{0}=20cm}

Hence 1 is correct option

 

Posted by

rishi.raj

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