A) 25 cm
B) 50 cm
C) 100 cm
D) 10 cm
Correct Answer: B
Solution :
Power (P) of a lens is inverse of focal length \[(f)\] (in metres). \[\therefore \] \[P=\frac{1}{f}\] \[\Rightarrow \] \[f=\frac{1}{P}=-\frac{100}{2}=-50\,cm\] Since, focal length is negative it is a concave lens, from lens formula, we have \[\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\] where v is image distance, u is object distance. Given, \[v=-25\text{ }cm,\text{ }f=-50\text{ }cm\] \[-\frac{1}{50}=-\frac{1}{25}-\frac{1}{u}\] \[\therefore \] \[\frac{1}{u}=-\frac{1}{25}+\frac{1}{50}=\frac{-2+1}{50}=-\frac{1}{50}\] \[\Rightarrow \] \[u=-\text{ }50\text{ }cm\] \[\therefore \] Object distance is 50 cm.You need to login to perform this action.
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