COMEDK · Physics · 25. Ray Optics
In the normal adjustment of an astronomical telescope, the objective and eyepiece are \(32 \mathrm{~cm}\) apart. If the magnifying power of the telescope is 7, find the focal lengths of the objective and eyepiece.
- A \(f_o=28 \mathrm{~cm}\) and \(f_e=7 \mathrm{~cm}\)
- B \(\mathrm{f}_{\mathrm{e}}=28 \mathrm{~cm}\) and \(\mathrm{f}_{\mathrm{o}}=4 \mathrm{~cm}\)
- C \(\mathrm{f}_{\mathrm{o}}=7 \mathrm{~cm}\) and \(\mathrm{f}_{\mathrm{e}}=28 \mathrm{~cm}\)
- D \(f_o=28 \mathrm{~cm}\) and \(f_e=4 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(D) \(f_o=28 \mathrm{~cm}\) and \(f_e=4 \mathrm{~cm}\)
Step-by-step Solution
Detailed explanation
In the normal adjustment of an astronomical telescope, the distance between the objective and the eyepiece is given by \(L = f_o + f_e\). Given \(L = 32 \text{ cm}\), we have \(f_o + f_e = 32\). The magnifying power \(M\) of an astronomical telescope in normal adjustment is…
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