A) Divergent in P, convergent in Q
B) Convergent in P, divergent in Q
C) Convergent in both
D) Divergent in both
Correct Answer: C
Solution :
For arrangement P, \[\frac{1}{{{F}_{eq}}}=\frac{1}{2R}+\frac{1}{2R}-\frac{2}{2R}=\frac{1}{3R}\] (R = radius of curvature) For arrangement Q, \[\frac{1}{{{F}_{eq}}}=\frac{1}{2R}+\frac{1}{2R}-\frac{1}{3R}=\frac{2}{3R}\] So equivalent focal length for arrangement P is greater than equivalent focal length for arrangement Q and of the same sign i.e., positive.You need to login to perform this action.
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