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(B) In normal adjustment,the angular magnification $m$ is given by $m = \frac{f_o}{f_e}$.Given $m = 10$ and $f_e = 5 \ cm$,we have $10 = \frac{f_o}{5}

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) In normal adjustment,the angular magnification $m$ is given by $m = \frac{f_o}{f_e}$.Given $m = 10$ and $f_e = 5 \ cm$,we have $10 = \frac{f_o}{5}$,which gives $f_o = 50 \ cm$.When the final image is formed at the least distance of distinct vision $D = 25 \ cm$,the angular magnification $m'$ is given by the formula $m' = \frac{f_o}{f_e} \left( 1 + \frac{f_e}{D} ight)$.Substituting the values: $m' = \frac{50}{5} \left( 1 + \frac{5}{25} ight)$.$m' = 10 	imes (1 + 0.2) = 10 	imes 1.2 = 12$.Therefore,the angular magnification is $12$.

An astronomical telescope has an eyepiece of focal length $5 \ cm$. The angular magnification in normal adjustment is $10$. When the final image is formed at the least distance of distinct vision $(25 \ cm)$ from the eyepiece,then the angular magnification will be:

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