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