A) + 1.5 D
B) \[-1.5\text{ }D\]
C) + 6.67D
D) \[-6.67D\]
Correct Answer: B
Solution :
Power\[=\frac{1}{focal\text{ }length}\] 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. Now, \[\frac{1}{F}=\frac{1}{0.4}+\frac{1}{(-0.25)}\] \[\frac{1}{F}=-1.5\] \[\therefore \] Power \[P=-1.5D\]
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