A) the focal length of the objective
B) the focal length of the eyepiece
C) aperture of the objective
D) aperture of the eyepiece
Correct Answer: A
Solution :
Magnifying power of a telescope is \[\text{M =}\frac{\text{angle subteded by final image at eye}}{\text{angle subtended by the object at eye}}\] When final image is formed at infinity, and \[{{f}_{o}}\] is focal length of objective, \[{{f}_{e}}\] is focal length of eye-piece then magnifying power is given by \[M=\frac{{{f}_{o}}}{{{f}_{e}}}\] Since, \[M\propto {{f}_{o}},\]therefore by increasing focal length of objective we can increase magnifying power.
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