COMEDK · Physics · 25. Ray Optics
The magnifying power of a telescope is 9 . When it is adjusted for parallel rays, the distance between the objective and eyepiece is \(20 \mathrm{~cm}\). The focal length of lenses are
- A \(10 \mathrm{~cm}, 10 \mathrm{~cm}\)
- B \(15 \mathrm{~cm}, 5 \mathrm{~cm}\)
- C \(18 \mathrm{~cm}, 2 \mathrm{~cm}\)
- D \(11 \mathrm{~cm}, 9 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(C) \(18 \mathrm{~cm}, 2 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
For a telescope adjusted for parallel rays, the magnifying power \(M\) is given by \(M = \dfrac{f_o}{f_e}\), where \(f_o\) is the focal length of the objective lens and \(f_e\) is the focal length of the eyepiece. Given \(M = 9\), we have \(\dfrac{f_o}{f_e} = 9\), which implies…
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