The magnifying power of a telescope is $9$ . When it is adjusted for parallel rays, the distance between the objective and the eye piece is found to be $20 c m$ . The focal length of lenses are

18 c m, 2 c m

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11 c m, 9 c m

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10 c m, 10 c m

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15 c m, 5 c m

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Solution

The correct option is B $18 c m, 2 c m$
$\frac{f_{0}}{f_{e}} = 9$ , $∴ f_{0} = 9 f_{e}$
Also $f_{0} + f_{e} = 20$ ( $∵$ final image is at infinity)
$9 f_{e} + f_{e} = 20, f_{e} = 2 c m, ∴ f_{0} = 18 c m$

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The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and the eye piece is found to be 20 cm. The focal length of lenses are

The correct option is B 18 cm,2 cmf0fe=9, ∴f0=9feAlso f0+fe=20 (∵ final image is at infinity)9fe+fe=20,fe=2cm,∴f0=18cm

The magnifying power of a telescope is $9$ . When it is adjusted for parallel rays, the distance between the objective and the eye piece is found to be $20 c m$ . The focal length of lenses are

The correct option is B $18 c m, 2 c m$
$\frac{f_{0}}{f_{e}} = 9$ , $∴ f_{0} = 9 f_{e}$
Also $f_{0} + f_{e} = 20$ ( $∵$ final image is at infinity)
$9 f_{e} + f_{e} = 20, f_{e} = 2 c m, ∴ f_{0} = 18 c m$