A) 1.66 D
B) 4.00 D
C) -1.00 D
D) - 3.75 D
Correct Answer: D
Solution :
\[\frac{1}{F}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}\]\[\Rightarrow \frac{1}{80}=\frac{1}{20}+\frac{1}{{{f}_{2}}}\] \[\Rightarrow {{f}_{2}}=-\frac{80}{3}cm\] \Power of second lens \[{{P}_{2}}=\frac{100}{{{f}_{2}}}=\frac{100}{-80/3}=-3.75\ D\]
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