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A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece? |
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Answer» Solution :Here, `f_(0) = 144 CM` and `f_(e) = 6.0 cm` `therefore` Magnifying power `=-f_(0)/f_(e) =(-144)/6 = -24` and separation between the two lenses `=f_(0) + f_(e) = 144 + 6 = 150 cm` or `1.5 m` |
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