The angular magnification of a refracting telescope is 40. When the object and f

Question

The angular magnification of a refracting telescope is 40. When the object and final image are both at infinity, the distance between the eyepiece and the objective is 143.5 cm. The telescope is used to view a distant radio tower. The real image of the tower, formed by the objective, is 6.0 mm in height. The focal point of the eyepiece is positioned at the real image. In the situation above, the angle subtended by the final image of the tower is closest to:

a) .23 rad
b) .21 rad
c) .17 rad
d) .19 rad
e) .15 rad
The correct answer is c) .17 rad, but I want to understand how it was calculated. Thank you!

Explanation / Answer

Use f(objective)=L/M so 143.5 cm/40 you get 0.035875 Then, plug that into theta= h/f(objective) 6.0 mm/ 0.035875= 0.167

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