A) \[+0.65D\]
B) \[-\,0.65D\]
C) \[+0.75D\]
D) \[-\,0.75D\]
Correct Answer: D
Solution :
Key Idea: The power of a thin lens is equal to the reciprocal of its focal length (f) measured in metres. Power of the combination is given by \[P=\frac{1}{f(metre)}\] Combined focal length is \[\frac{1}{f}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}\] Given,\[{{f}_{1}}=80\,cm,\,{{f}_{2}}=-50\,cm\](concave) \[\therefore \] \[\frac{1}{f}=\frac{1}{80}-\frac{1}{50}\] \[=-\frac{30}{4000}\] \[\Rightarrow \] \[f=-\frac{4000}{30}cm\] \[\therefore \] \[power=-\frac{100}{400/3}\] \[=\frac{-3}{4}=-0.75D\] Note: As focal length of combination is negative, hence combination behaves like a diverging lens.
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