NEET 2024 Physics Question 36 Solution
Question
A small telescope has an objective of focal length 140 cm and an eye piece of focal length 5.0 cm. The magnifying power of telescope for viewing a distant object is:
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A small telescope has an objective of focal length 140 cm and an eye piece of focal length 5.0 cm. The magnifying power of telescope for viewing a distant object is:
Magnifying power m = f_o / f_e = 140 / 5.0 = 28!
For an astronomical refracting telescope viewing a distant celestial object under normal adjustment (relaxed eye, final image at infinity), the angular magnification m is given by the ratio of the focal length of the objective lens to that of the eyepiece lens:
Substituting f_o = 140 cm and f_e = 5.0 cm:
Diagnose every choice: Know exactly why the right option works and why distractors fail.
Arbitrary incorrect value.
Incorrect division.
Incorrect.
Exact quotient: 140 / 5 = 28.
Students who confuse microscope and telescope formulas sometimes divide f_e by f_o, getting a fraction < 1.
"The magnifying power of an astronomical telescope in normal adjustment is m = f_o / f_e."
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