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Question

(D) The magnifying power $M$ of an astronomical telescope in normal adjustment is given by the ratio of the focal length of the objective lens $(f_o)$

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(D) The magnifying power $M$ of an astronomical telescope in normal adjustment is given by the ratio of the focal length of the objective lens $(f_o)$ to the focal length of the eye-piece $(f_e)$: $M = \frac{f_o}{f_e}$. Since power $P = \frac{1}{f}$ (in meters),we have $f = \frac{1}{P}$. Substituting this into the formula,we get $M = \frac{1/P_o}{1/P_e} = \frac{P_e}{P_o}$. Given: Power of objective $P_o = 0.5\, D$ and power of eye-piece $P_e = 20\, D$. Therefore,$M = \frac{20}{0.5} = 40$.

An astronomical telescope has objective and eye-piece lenses of powers $0.5\, D$ and $20\, D$ respectively. Its magnifying power will be:

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