A) \[+250\,cm\]
B) \[+155\text{ }cm\]
C) 25 cm
D) \[+15\text{ }cm\]
Correct Answer: D
Solution :
Magnifying power of an astronomical telescope is given by \[\left| m \right|\,\,=\,\,\frac{{{f}_{0}}}{{{f}_{e}}}\left( 1+\frac{{{f}_{e}}}{D} \right)\] where \[{{f}_{0}}\] is focal length of objective, \[{{f}_{e}}\] is focal length of eyepiece and D is least distance of distinct vision. From this formula we observe that \[Magnifying\,\,power\,\,=\,\,\frac{1}{focal\,\,length\,\,of\,\,eye-peice}\] Hence, to produce the largest magnification the focal length of eye-piece must be smallest. Hence, \[{{f}_{e}}=\,\,+15\,cm\]
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