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Geometrical Optics question

2012 · Q143
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Geometrical Optics question

2012 · Q143

NEETPhysicsGeometrical OpticsMCQ+4 / −1
The magnifying power of a telescope is 9. When it is adjusted for parallel rays the distance between the objective and eyepiece is 20 cm. The focal length of lenses are
  1. A
    10 cm, 10 cm
  2. B
    15 cm, 5 cm
  3. C
    18 cm, 2 cm
  4. D
    11 cm, 9 cm
View written solutionFree

Correct answer: C

Magnifying power, m = f0fe{{{f_0}} \over {{f_e}}}fe​f0​​ = 9 ..........(1)

where fo and fe are the focal lengths of the objective and eyepiece respectively

Also,fo + fe = 20 cm ...(2)

On solving (1) and (2), we get

fo = 18 cm, fe = 2 cm

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