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The focal length of the objective of an astronomical telescope is 75 cm and that of the eyepiece is 5 cm . If the final image is formed at the least distance of distanct of distinct vision from the eye, calculate the magnifying power of the telescope. |
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Answer» SOLUTION :DATA supplied , `f_(0) = 75 cm , "" f_(e) = 25 cm , "" D = 25 ` cm m = `- (f_(0))/(f_(e)) ( 1 + (f_(e))/(D) ) = (-75)/(5) (1 + (5)/(25)) = -15 xx 1.2 = - 18` |
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