If tube length of the astronomical telescope is 105 cm and magnifying power is 20 for normal setting. The focal length of an objective is:

100 cm

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10 cm

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20 cm

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25 cm

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Solution

The correct option is A 100 cm
Tube length, $L = f_{o} + f_{e} = 105 c m$

Magnification, $m = \frac{f_{o}}{f_{e}} = 20$

$\Rightarrow f_{o} = 20 f_{e}$

$20 f_{e} + f_{e} = 105$

$\Rightarrow 21 f_{e} = 105$

$\Rightarrow f_{e} = \frac{105}{21}$

$\Rightarrow f_{e} = 5 c m$

$∴ f_{o} = 20 \times 5 = 100 c m$

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If tube length of the astronomical telescope is 105 cm and magnifying power is 20 for normal setting. The focal length of an objective is:

The correct option is A 100 cmTube length, L=fo+fe=105cmMagnification, m=fofe=20⇒fo=20fe20fe+fe=105⇒21fe=105⇒fe=10521⇒fe=5cm∴fo=20×5=100cm