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A small telescope has an objective lens of focal length 150 cm and eyepiece of focal length 5 cm. What is the magnifying power of the telescope for viewing distant obj ects in normal adjustment ? If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens ? |
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Answer» SOLUTION :Here, `f_(0) = 150 cm ` and `f_(e) = 5 cm` Magnifying power of telescope in normal adjustment `m=-f_(0)/f_(e) = -150/5 = -30` The -ve sign signifies that the image formed is an inverted image. Again height of tower H = 100 m and DISTANCE of tower d = 3 km = 3000 m. `therefore`Angle subtended by the tower at the objective lens a is given by `alpha = h/d = h^(.)/f_(0)`, where h. is the height of the image of the tower formed by the objective lens. `rArr h. = (h.f_(0))/d = ((100 m) xx (150 cm))/(3000 m) =((100 m) xx (1.5 m))/(3000 m) = 0.05` or `5.0 cm` |
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