A) ? 2 D and 0.33 D
B) 2 D and ? 0.33 D
C) ? 2 D and 3 D
D) 2 D and ? 3 D
Correct Answer: B
Solution :
For correcting the near point, required focal length \[f=\frac{50\times 25}{(50-25)}=50\,cm\] So power \[P=\frac{100}{50}=+\,2\,D\] For correcting the far point, required focal length \[f=-\,(defected\ far\ point)=-\,3\,m\] \[\therefore P=-\frac{1}{3}D=-\,0.33\,D\]You need to login to perform this action.
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