A) 40
B) 80
C) 50
D) 0.01
Correct Answer: A
Solution :
In relaxed eye state of telescope or when final image is formed at infinity, the magnification of telescope is given by \[\left| M \right|=\frac{{{f}_{0}}}{{{f}_{e}}}\] where \[{{f}_{0}}=\] focal length of objective = 200 cm \[{{f}_{e}}=\]focal length of eye-piece = 5 cm Hence, \[\left| M \right|=\frac{200}{5}=40\] Note: The aperture of the objective lens is kept large so that more and more rays coming from the heavenly body may enter the telescope and a bright image is formed by the objective lens. The aperture of the eye lens is kept comparatively small so that all the rays may enter the eye.
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