COMEDK · Physics · 25. Ray Optics
In the normal adjustment of an astronomical telescope, the objective and eyepiece are 36 cm apart. If the magnifying power of the telescope is 8, find the focal lengths of the objective and eyepiece.
- A \(\mathrm{F}_{\mathrm{o}}=28 \mathrm{~cm}, \mathrm{F}_{\mathrm{e}}=7 \mathrm{~cm}\)
- B \(\mathrm{F}_{\mathrm{o}}=4 \mathrm{~cm}, \mathrm{F}_{\mathrm{e}}=32 \mathrm{~cm}\)
- C \(\mathrm{F}_{\mathrm{o}}=28 \mathrm{~cm}, \mathrm{F}_{\mathrm{e}}=4 \mathrm{~cm}\)
- D \(\mathrm{F}_{\mathrm{o}}=32 \mathrm{~cm}, \mathrm{F}_{\mathrm{e}}=4 \mathrm{~cm}\)
Answer & Solution
Correct Answer
(D) \(\mathrm{F}_{\mathrm{o}}=32 \mathrm{~cm}, \mathrm{F}_{\mathrm{e}}=4 \mathrm{~cm}\)
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 \(L = 36 \text{ cm}\), we have \(f_o + f_e = 36\). The magnifying power \(M\) of an astronomical telescope in normal adjustment is…
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