A) 3D
B) \[-3D\]
C) \[~-1.5\text{ }D\]
D) \[4-1.5\text{ }D\]
Correct Answer: D
Solution :
Key Idea: \[Power=\frac{1}{focal\,length\,(m)}\] From lens formula \[\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\] Given, \[v=-100\,cm,\,u=-40\,cm\] \[\therefore \] \[\frac{1}{f}=-\frac{1}{100}-\frac{1}{-40}\] \[\Rightarrow \] \[\frac{1}{f}=-\frac{1}{100}+\frac{1}{40}\] \[\Rightarrow \]\[\frac{1}{f}=\frac{-1+2.5}{100}=\frac{1.5}{100}\] Hence, power of lens is \[P=\frac{100}{f(cm)}=100\times \frac{1.5}{100}=1.5\,D\]
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