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 \[|M|\,\,=\frac{{{f}_{o}}}{{{f}_{e}}}\] where \[{{f}_{0}}=\] 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 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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