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\[M\text{=}\frac{\text{angle}\,\,\text{subtended}\,\,\text{by}\,\,\text{final}\,\,\text{image}\,\,\text{at}\,\,\text{eye}}{\text{angle}\,\,\text{subtended}\,\,\text{by}\,\,\text{the}\,\,\text{object}\,\,\text{at}\,\,\text{eye}}\] When final image is formed at infinity, and \[{{f}_{o}}\] is focal length of objective, /g 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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