Magnifying power of an astronomical telescope is $$3$$ for normal vision and when the final image is formed at least distance of distinct vision, the magnifying power is $$9$$. Then tube length of the telescope for normal vision is [Least distance for distinct vision $$=$$ $$25$$ cm]

Question

Infinity Learn · Accepted Answer

For normal vision, $$f0fe=3When$$ the image is formed at least distance of distinct vision, $$f0fe1+feD=9$$⇒$$31+fe25=9$$⇒$$fe=50$$ cm∴$$f0=3fe=150$$ cm∴For normal vision, $$L=f0+fe=150+50$$ $$cm=200$$ cm

Magnifying power of an astronomical telescope is 3 for normal vision and when the final image is formed at least distance of distinct vision, the magnifying power is 9. Then tube length of the telescope for normal vision is [Least distance for distinct vision = 25 cm]

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