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

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
$m = \frac{f_{0}}{f_{e}}; 20 = \frac{f_{0}}{f_{e}} f_{e} = \frac{f_{0}}{20} L = f_{0} + f_{e} 105 = f_{0} + f_{e} 105 = f_{0} + \frac{f_{0}}{20} 105 = (\frac{20 + 1}{20}) f_{0} f_{0} = \frac{105 \times 20}{21} f_{0} = 100 c m$

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

If the tube length of the astronomical telescope is 105 cm long and magnifying power is 20 for normal setting. Calculate the focal length of the objective. Given magnifying power(m)= $m = \frac{f_{0}}{f_{e}}$ .

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

The correct option is A 100 cmm=f0fe;20=f0fefe=f020L=f0+fe105=f0+fe105=f0+f020105=(20+120)f0f0=105×2021f0=100cm

The correct option is A 100 cm
$m = \frac{f_{0}}{f_{e}}; 20 = \frac{f_{0}}{f_{e}} f_{e} = \frac{f_{0}}{20} L = f_{0} + f_{e} 105 = f_{0} + f_{e} 105 = f_{0} + \frac{f_{0}}{20} 105 = (\frac{20 + 1}{20}) f_{0} f_{0} = \frac{105 \times 20}{21} f_{0} = 100 c m$