A telescope has an objective lens of focal le | vedclass

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(A) For a telescope in normal adjustment,the magnifying power $(M)$ is given by the ratio of the focal length of the objective lens $(f_o)$ to the foc

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(A) For a telescope in normal adjustment,the magnifying power $(M)$ is given by the ratio of the focal length of the objective lens $(f_o)$ to the focal length of the eyepiece $(f_e)$.Given:Focal length of objective lens,$f_o = 144\, cm$Focal length of eyepiece,$f_e = 6\, cm$The formula for magnifying power is:$M = -\frac{f_o}{f_e}$Substituting the values:$M = -\frac{144}{6} = -24$The magnitude of the magnifying power is $24$.

$A$ telescope has an objective lens of focal length $144\, cm$ and an eyepiece of focal length $6\, cm$. Find the magnifying power of the telescope for normal adjustment.

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