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Wave Optics

Question
CBSEENPH12039792
Wired Faculty App

An astronomical telescope has objective eyepiece of focal lengths 40 cm and 4 cm respectively. To view an object 200 cm away from the objective lens, the lenses must be separated by a distance,

  • 46.0 cm

  • 50.0 cm

  • 54.0 cm

  • 37.3 cm

Solution

C.

54.0 cm

According to a question,
Focal length of the objective lens, Fo = +40 cm
Focal length of eyepiece, Fe = 4 cm
Object distance for objective lens (uo) = -200 cm
Applying lens formula for objective lens,

space space space space space 1 over straight v minus 1 over u equals 1 over f rightwards double arrow space 1 over v space minus space fraction numerator 1 over denominator negative 200 end fraction space equals space 1 over 40 rightwards double arrow space 1 over v equals space 1 over 40 space minus 1 over 200 space space space space space space space space space space equals space fraction numerator 5 minus 1 over denominator 200 end fraction space space space space space space space space space equals space 4 over 200 rightwards double arrow space v space equals space 50 space c m
Image will be formed at the focus of a eyepiece lens.
So, for normal adjustment distance between objectives and eyepiece (length of tube) will be,
straight v space plus space straight F subscript straight e space equals space 50 space plus space 4 space equals space 54 space cm

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