The focal length of the objective and the eye | vedclass

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(B) Given:Focal length of objective,$f_o = 100\, cm$Focal length of eyepiece,$f_e = 5\, cm$Least distance of distinct vision,$D = 25\, cm$The formula

Vedclass · Accepted Answer

$24$

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(B) Given:Focal length of objective,$f_o = 100\, cm$Focal length of eyepiece,$f_e = 5\, cm$Least distance of distinct vision,$D = 25\, cm$The formula for the magnification of a telescope when the final image is formed at the least distance of distinct vision is given by:$m = \frac{f_o}{f_e} \left( 1 + \frac{f_e}{D} ight)$Substituting the values:$m = \frac{100}{5} \left( 1 + \frac{5}{25} ight)$$m = 20 \left( 1 + 0.2 ight)$$m = 20 	imes 1.2 = 24$Therefore,the magnification of the telescope is $24$.

The focal length of the objective and the eyepiece of a telescope are $100\, cm$ and $5\, cm$ respectively. If the final image is formed at the least distance of distinct vision,the magnification of the telescope is:

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