The focal length of the objective lens of a telescope is 50 cm. If the magnification of the telescope is 25, then the focal length of the eye-piece is
- A.12.5 cm
- B.5 cm
- C.2 cm
- D.10 cm
Correct answer
Explanation
The correct answer is C, 2 cm. For an astronomical telescope in normal adjustment the magnifying power equals the focal length of the objective divided by that of the eyepiece, so the eyepiece focal length is 50 ÷ 25 = 2 cm. A refracting telescope uses an objective lens of long focal length and large aperture to gather light and form a real image of a distant object, and an eyepiece of short focal length to magnify that image. In normal adjustment the final image is at infinity and the tube length is the sum of the two focal lengths, here 52 cm. This is why a high-power telescope pairs a long objective with a very short eyepiece. Option A is wrong because 12.5 cm would give a magnification of only 4. Option B is wrong because 5 cm would give a magnification of 10. Option D is wrong because 10 cm would give a magnification of 5. Exam tip: telescope M = fo ÷ fe, length = fo + fe; objective long, eyepiece short.
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