The diameter of the moon is $$3.5$$ × $$103$$ km and its distance from earth is $$3.8$$ × $$105$$ km. When seen by a telescope having focal lengths of the objective and the eye piece as $$40$$ mm and $$10$$ mm respectively, the angle subtended by the image of the moon at the eyes for normal vision will be approximately

Question

The diameter of the moon is $$3.5$$  × $$103$$  km and its distance from earth is $$3.8$$  × $$105$$  km. When seen by a telescope having focal lengths of the objective and the eye piece as $$40$$ mm and $$10$$ mm respectively, the angle subtended by the image of the moon at the eyes for normal vision will be approximately

Infinity Learn · Accepted Answer

$$(1)Angular$$ magnification $$=$$ βα$$=f0feNow$$  α  $$=$$  $$3.5$$  ×  $$1033.8$$  ×  $$105rad=9.210$$  ×  $$10$$−$$3rad$$.∴   β  $$=9.210$$   ×  $$10$$−$$3$$  ×  $$4010$$  rad  ≃ $$20$$

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The diameter of the moon is 3.5 × 103 km and its distance from earth is 3.8 × 105 km. When seen by a telescope having focal lengths of the objective and the eye piece as 40 mm and 10 mm respectively, the angle subtended by the image of the moon at the eyes for normal vision will be approximately

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