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An astronomical telescope has objective and eye-piece lenses of powers 0.5 D and 20 D respectively. What will be its magnifying power?

Asked in GUJCET 2009 · Telescopes

• (1) 10
• (2) 20
• (3) 30
• (4) 40

Step-by-step solution

Why the other options are wrong

• (1) 10 is the product 0.5× 20 of the two powers. The magnifying power is a ratio, m=(fₒ)/(fₑ)=(Pₑ)/(Pₒ)=(20)/(0.5)=40.
• (2) 20 is the eye-piece's power read straight off as the magnification. The eye-piece alone fixes nothing; it is the long objective focal length divided by the short eye-piece one that magnifies.
• (3) No combination of 0.5 D and 20 D gives 30. Sanity check: fₒ=1/(0.5)=2 m and fₑ=1/(20)=0.05 m, so m=2/(0.05)=40.

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