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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 object in normal adjustment? If this telescope is used to view a 100 m tall Also 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 :If the telescope is in normal adjustment, i.e., the FINAL image is at infinity. Formula : ` M = (f_o)/(f_e)` since ` f_o = 150 cm , f_e = 5cm` ` therefore M = 150/5 = 30 ` If tall tower is at distance 3 KM from the objective lens of focal length 150 cm. It will form a image at distance `v_o`So, ` (1)/(f_o) = (1)/(v_o) - (1)/(u_o)` ` (1)/(150cm) = (1)/(v_o) - (1)/(-3km)` ` (1)/(v_o) = (1.5 m) - (1)/(3000m)` ` v_o = (3000 XX 1.5 )/(3000 - 1.5)` `= (4500)/(2998.5) = 1.5 m` MAGNIFICATION `m_o =1/O = (h_1)/(h_o) = (v_o)/(u_o)` ` (h_i)/(100m) = (1.5 m)/(3km) = (1.5)/(3000) ` ` h_i= (1.5 xx 100)/(3000) = 1/20 m` ` h_i = 0.05 m. ` |
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