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प्रश्न
The magnifying power of an astronomical telescope in normal adjustment is 2.9 and the objective and the eyepiece are separated by a distance of 150 cm. Find the focal lengths of the two lenses.
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उत्तर
Given: D = 150 cm, m = 2.9
fo = ?
Since, `m = f_o/f_e`
D = `f_o + f_e`
`2.9 = f_o/f_e`
`f_o = f_e xx 2.9` .....(i)
`f_o + f_e = 150` cm
From equation (i),
`2.9 f_e + f_e = 150`
`3.9f_e = 150`
fe = 38.46 cm
fo = 2.9 × 38.46
fo = 111.54 cm
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संबंधित प्रश्न
A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when
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- the final image is formed at the least distance of distinct vision (25 cm)?
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2) resolving power
Draw a ray diagram showing the image formation of a distant object by a refracting telescope ?
"A telescope resolves whereas a microscope magnifies." Justify this statement ?
Define its magnifying power and write the expression for it?
A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece lens of focal length 1.0 cm is used, find the angular magnification of the telescope. If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.42 × 106 m and the radius of the lunar orbit is 3.8 × 108 m.
Draw a labelled ray diagram showing the formation of an image by a refracting telescope when the final image lies at infinity.
Define the term ‘resolving power of a telescope’.
Draw a labelled ray diagram showing the image formation by a refracting telescope. Define its magnifying power.
