Question:
The distance between the objective and eyepiece of an astronomical telescope when the final image forms at infinity is 62 cm. If the magnification of the telescope is 30, the focal lengths of the objective and eyepiece respectively are:
The distance between the objective and eyepiece of an astronomical telescope when the final image forms at infinity is 62 cm. If the magnification of the telescope is 30, the focal lengths of the objective and eyepiece respectively are:
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Astronomical telescope: $L = f_o + f_e$ when final image at infinity.
Angular magnification $M = f_o / f_e$.
Solve two equations simultaneously to get $f_o$ and $f_e$.
Angular magnification $M = f_o / f_e$.
Solve two equations simultaneously to get $f_o$ and $f_e$.
Updated On: Mar 6, 2026
- 31 cm, 2 cm
- 2 cm, 31 cm
- 2 cm, 60 cm
- 60 cm, 2 cm
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The Correct Option is A
Solution and Explanation
• For astronomical telescope: $L = f_o + f_e = 62\ \text{cm}$, magnification $M = f_o / f_e = 30$.
• Let $f_o = 30 f_e \Rightarrow f_o + f_e = 30 f_e + f_e = 31 f_e = 62 \Rightarrow f_e = 2\ \text{cm}$, $f_o = 30 \cdot 2 = 60\ \text{cm}$. Wait, distance between lenses = $f_o + f_e = 31*2 = 62$, so yes $f_o = 60\ \text{cm}$, $f_e = 2\ \text{cm}$ which matches option (1).
• Hence, focal lengths are 60 cm and 2 cm.
• Let $f_o = 30 f_e \Rightarrow f_o + f_e = 30 f_e + f_e = 31 f_e = 62 \Rightarrow f_e = 2\ \text{cm}$, $f_o = 30 \cdot 2 = 60\ \text{cm}$. Wait, distance between lenses = $f_o + f_e = 31*2 = 62$, so yes $f_o = 60\ \text{cm}$, $f_e = 2\ \text{cm}$ which matches option (1).
• Hence, focal lengths are 60 cm and 2 cm.
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