A) 6.0 D
B) 1.5 D
C) -6.0 D
D) -1.5 D
Correct Answer: D
Solution :
Key Idea: Power of a lens is equal to inverse of its focal length. The power of a thin lens is equal to the reciprocal of its focal length \[(f)\] measured in metres \[P=\frac{1}{f(metres)}\] Given, focal length of convex lens \[{{f}_{1}}=40\,\text{cm}\]and of concave lens \[{{f}_{2}}=-25\,cm\] \[\therefore \] Focal length of combination \[\frac{1}{F}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}=\frac{1}{40}-\frac{1}{25}=-\frac{3}{200}\] \[\Rightarrow \] \[F=-\frac{200}{3}=-66.7\,cm\] Power of spectacles is \[=\frac{100}{F}=\frac{100}{-66.7}=-1.5D\]
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