The focal length of an objective of a telesco | vedclass

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(A) For the complete use of the objective lens,the diameter of the exit pupil of the telescope must match the diameter of the human eye's pupil. The e

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(A) For the complete use of the objective lens,the diameter of the exit pupil of the telescope must match the diameter of the human eye's pupil. The exit pupil is the image of the objective formed by the eyepiece.The magnification $M$ of a telescope is given by the ratio of the focal lengths of the objective $(f_o)$ and the eyepiece $(f_e)$:$M = \frac{f_o}{f_e}$Also,the magnification is the ratio of the diameter of the objective $(D)$ to the diameter of the exit pupil $(d)$:$M = \frac{D}{d}$Equating the two expressions for magnification:$\frac{f_o}{f_e} = \frac{D}{d}$Given:$f_o = 3 \ m = 300 \ cm$$D = 15 \ cm$$d = 3 \ mm = 0.3 \ cm$Substituting the values:$\frac{300}{f_e} = \frac{15}{0.3}$$\frac{300}{f_e} = 50$$f_e = \frac{300}{50} = 6 \ cm$

The focal length of an objective of a telescope is $3 \ m$ and its diameter is $15 \ cm$. Assuming that for a normal eye,the diameter of the pupil is $3 \ mm$ for its complete use,the focal length of the eyepiece must be: (in $cm$)

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