For a telescope,the focal length of the objec | vedclass

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(B) Given: Focal length of objective lens $f_o = 15 \ cm = 150 \ mm$. Focal length of eyepiece $f_e = 10 \ mm = 1 \ cm$. Tube length $L = 16 \ cm$. Fo

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$15$

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(B) Given: Focal length of objective lens $f_o = 15 \ cm = 150 \ mm$. Focal length of eyepiece $f_e = 10 \ mm = 1 \ cm$. Tube length $L = 16 \ cm$. For a telescope in normal adjustment,the tube length $L = f_o + f_e$. Here,$f_o + f_e = 15 \ cm + 1 \ cm = 16 \ cm$,which matches the given tube length. Thus,the telescope is in normal adjustment. The magnification $M$ for a telescope in normal adjustment is given by $M = \frac{f_o}{f_e}$. $M = \frac{15 \ cm}{1 \ cm} = 15$. Therefore,the magnification is $15$.

For a telescope,the focal length of the objective lens is $15 \ cm$ and the focal length of the eyepiece is $10 \ mm$. If the tube length is $16 \ cm$,find the magnification.

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