A) +1D
B) +9D
C) -1D
D) -9D
Correct Answer: B
Solution :
Power of first lens \[{{P}_{1}}=\frac{100}{{{f}_{1}}}=5D\] Power of second lens \[{{P}_{1}}=\frac{100}{{{f}_{1}}}=\frac{100}{20}\,=5D\] Total power \[P={{P}_{1}}+{{P}_{2}}=5+4=9\,D\]
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