AP EAMCET · PHYSICS · Ray Optics
The magnifying power of a telescope with tube length 60 \(\mathrm{cm}\) is 5 . Then the focal length of its eye piece is
- A \(20 \mathrm{~cm}\)
- B \(40 \mathrm{~cm}\)
- C \(30 \mathrm{~cm}\)
- D \(10 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(D) \(10 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
The magnifying power of a telescope is given by \(\begin{aligned} & m=\frac{f_0}{f_e}=5 \\ & f_0=5 f_e \\ & f_0+f_e=60 \mathrm{~cm} \Rightarrow 5 f_e+f_e=60 \\ & 6 f_e=60 \Rightarrow f_e=10 \mathrm{~cm} \end{aligned}\) The focal length of its eye piece, \(f_e=10 \mathrm{~cm}\)
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