Ray Optics

JEE Main level, single correct, medium

277 of 295

Question 277 of 295

In normal adjustment, for a refracting telescope, the distance between objective and eye piece is 30cm30cm. The focal length of the objective, when the angular magnification of the telescope is 22, will be:

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Answer: (A) 20cm20cm

Let f0{f}_{0} be the focal length of objective and fe{f}_{e} is the focal length of eyepiece.
Given here, f0+fe=30{f}_{0}+{f}_{e}=30.
Angular magnification of telescope is given by m=f0fem=\dfrac{{f}_{0}}{{f}_{e}}.
So, 2=f0fe⇒f0=2fe2=\dfrac{{f}_{0}}{{f}_{e}}\Rightarrow {f}_{0}=2{f}_{e}
Then, we have f0+f02=30⇒3f02=30{f}_{0}+\dfrac{{f}_{0}}{2}=30\Rightarrow \dfrac{3{f}_{0}}{2}=30
Or, f0=20cm{f}_{0}=20cm.

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