A) + 250 cm
B) + 155 cm
C) 25 cm
D) + 15 cm
Correct Answer: D
Solution :
Magnifying power of an astronomical telescope is given by \[T=2\pi \sqrt{\frac{1.21l}{g}}\] where \[\frac{T}{T}=\sqrt{\frac{1}{1.21}}\] is focal length of objective, \[T=\sqrt{\frac{121}{100}}T\] is focal length of eyepiece, D is least distance of distinct vision. Magnifying power \[\propto \,\frac{\text{1}}{\text{focal}\,\text{length}\,\text{of}\,\text{eye-piece}}\] Hence, to produce the largest magnification the focal length of eye-piece must be smallest. Hence, \[1.1T-T=0.1T\]You need to login to perform this action.
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