$A$ small telescope has an objective lens of focal length $144\;cm$ and an eyepiece of focal length $6.0\;cm$. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?
(N/A) Given:
Focal length of the objective lens,$f_{o} = 144\;cm$
Focal length of the eyepiece,$f_{e} = 6.0\;cm$
The magnifying power $(m)$ of the telescope in normal adjustment is given by the formula:
$m = \frac{f_{o}}{f_{e}}$
$m = \frac{144}{6} = 24$
The separation between the objective lens and the eyepiece for normal adjustment is given by:
$L = f_{o} + f_{e}$
$L = 144 + 6 = 150\;cm$
Thus,the magnifying power of the telescope is $24$ and the separation between the objective lens and the eyepiece is $150\;cm$.
Focal length of the objective lens,$f_{o} = 144\;cm$
Focal length of the eyepiece,$f_{e} = 6.0\;cm$
The magnifying power $(m)$ of the telescope in normal adjustment is given by the formula:
$m = \frac{f_{o}}{f_{e}}$
$m = \frac{144}{6} = 24$
The separation between the objective lens and the eyepiece for normal adjustment is given by:
$L = f_{o} + f_{e}$
$L = 144 + 6 = 150\;cm$
Thus,the magnifying power of the telescope is $24$ and the separation between the objective lens and the eyepiece is $150\;cm$.
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