A) \[3D\]
B) \[-3D\]
C) \[-1.5D\]
D) \[+1.5D\]
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, \[v=-1000\,\,cm,\,\,u=-40\,\,cm\] \[\therefore \] \[\frac{1}{f}=-\frac{1}{100}-\frac{1}{-40}\] \[\Rightarrow \] \[\frac{1}{f}=-\frac{1}{100}+\frac{1}{40}\] \[\Rightarrow \] \[\frac{1}{f}=\frac{-1+2.5}{100}=\frac{1.5}{100}\] Hence, power of lens is \[P=\frac{100}{f(cm)}=100\times \frac{1.5}{100}=1.5\,\,D\]
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