The magnifying power of a telescope with tube 60 cm is 5. What is the focal length of its eye piece ?
Solution
For telescope
<br><br>Tube length (L) = f<sub>0</sub> + f<sub>e</sub>
<br><br>and magnification (m) = ${{{f_0}} \over {{f_e}}}$ = 5
<br><br>where f<sub>o</sub> and f<sub>e</sub> are focal length of objective
and eyepiece.
<br><br>$\therefore$ f<sub>0</sub> + f<sub>e</sub> = 60
<br><br>$\therefore$ f<sub>o</sub> = 50 cm
<br><br>f<sub>e</sub> = 10 cm
About this question
Subject: Physics · Chapter: Optics · Topic: Lenses and Optical Instruments
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