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(a) For the telescope described in Exercise 9.28(a), what is the separation between the objective lens and the eyepiece ? (b) 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 ? ( c) What is the height of the final image of the tower if it is formed at 25 cm ? |
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Answer» Solution :(a) SEPARATION between the objective lens and eyepiece = `f_0+f_e= 140 + 5 = 145 cm` or 1.45 m. (B) Angle subtended at the TELESCOPE objective by the 100 m tall tower situated 3 km (= 3000 m) AWAY `alpha = 100/(3000 = 1/30` rad If image formed by objective lens in its focal plane be of height h, then `alpha = h/f_(0)` or `h =alpha f_(0) = 1/30 xx 140 cm = 4.67 cm` (c) When final image is formed at d = 25 cm, the magnification produced by the eyepiece: `m_( e) = 1+ D/f_(e) = 1+ 25/5 = 6` `therefore`Height of final image of the tower `h. = hm_(e) = 4.67 xx 6 = 28 cm` |
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