InterviewSolution
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InterviewSolution
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A small telescope has an objective lens of focal length `140 cm` and eye piece of focal length `5.0 cm`. What is the magnifying power of telescope for viewing distant objects when (a) the telescope is in normal adjustment (i.e. when the image is at infinity) (b) the final image is formed at the least distance of distinct vision `(25 cm)`. |
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Answer» Here, `f_0 = 140 cm, f_e = 5.0 cm` Magnifying power = ? (a) In normal adjustment, Magnifying power `= (f_0)/(-f_e) = (140)/(-5) = - 28` (b) When final image is at the least distance of distinct vision, Magnifying power `= (-f_0)/(f_e) (1 + (f_e)/(d)) = (140)/(-5) (1 + (5)/(25)) = -33.6`. |
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