A) + 0.65 D
B) -0.65 D
C) + 0.75 D
D) -0.75 D
Correct Answer: D
Solution :
\[{{f}_{1}}=+\,80\,cm,\] \[{{f}_{2}}=-\,50\,cm.\] Resultant focal length \[F=\frac{{{f}_{1}}{{f}_{2}}}{{{f}_{1}}+{{f}_{2}}}\] \[=\frac{80\times -\,50}{80-50}=\frac{-\,4000}{30}\] \[=-\,\frac{4000}{3}cm.\] Resultant power\[P=\frac{100}{F}=\frac{100}{-\,400/3}\] \[=-\frac{3}{4}\] \[=-\,0.75D\]
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