Magnification produced by an astronomical tel | vedclass

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(A) For normal adjustment,the magnification $m = f_o / f_e = 10$,which implies $f_o = 10 f_e$.The length of the telescope is $L = f_o + f_e = 1.1 \, m

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(A) For normal adjustment,the magnification $m = f_o / f_e = 10$,which implies $f_o = 10 f_e$.The length of the telescope is $L = f_o + f_e = 1.1 \, m = 110 \, cm$.Substituting $f_o = 10 f_e$ into the length equation: $10 f_e + f_e = 110 \, cm \implies 11 f_e = 110 \, cm \implies f_e = 10 \, cm$.Then,$f_o = 10 	imes 10 \, cm = 100 \, cm$.When the image is formed at the least distance of distinct vision $(D = 25 \, cm)$,the magnification is given by $m' = f_o / f_e 	imes (1 + f_e / D)$.$m' = 10 	imes (1 + 10 / 25) = 10 	imes (1 + 0.4) = 10 	imes 1.4 = 14$.

Magnification produced by an astronomical telescope for normal adjustment is $10$ and the length of the telescope is $1.1 \, m$. The magnification when the image is formed at the least distance of distinct vision $(D = 25 \, cm)$ is

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