A) -0.5 D
B) + 0.5 D
C) - 0.625 D
D) + 0.625 D
Correct Answer: C
Solution :
Total power \[P={{P}_{1}}+{{P}_{2}}=11-6=5D\] Also \[\frac{{{f}_{l}}}{{{f}_{a}}}=\frac{{{(}_{a}}{{\mu }_{g}}-1)}{{{(}_{l}}{{\mu }_{g}}-1)}\Rightarrow \frac{{{P}_{a}}}{{{P}_{l}}}=\frac{{{(}_{a}}{{\mu }_{g}}-1)}{{{(}_{l}}{{\mu }_{g}}-1)}\] \[\Rightarrow \]\[\frac{5}{{{P}_{l}}}=\frac{(1.5-1)}{(1.5/1.6-1)}\Rightarrow {{P}_{l}}=-0.625\,D\]
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