A) +1.5 D
B) -1.5 D
C) 6.67 D
D) -6.67 D
Correct Answer: B
Solution :
\[\text{Power}\,=\frac{\text{1}}{\text{Focal}\,\text{length}}\] Focal length of combination of convex and concave lens is given by \[\frac{1}{F}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}\] where,\[{{f}_{1}}\]and\[{{f}_{2}}\]be the focal lengths of convex and concave lenses respectively \[\frac{1}{F}=\frac{1}{0.4}+\frac{1}{(-\,0.25)}\] \[\frac{1}{F}=-\,1.5\] \[P=-\,1.5\,D\]
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