A) At one of the foci, virtual and double its size
B) At 3f / 2, real and inverted
C) At 2f, virtual and erect
D) None of these
Correct Answer: A
Solution :
\[\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\] (Given \[u=\frac{-f}{2}\]) \[\Rightarrow \] \[\frac{1}{f}=\frac{1}{v}+\left( \frac{1}{f/2} \right)\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{2}{f}\] \[\Rightarrow \] \[\frac{1}{v}=\frac{-1}{f}\] and \[m=\frac{v}{u}=\frac{f}{f/2}=2\] So virtual at the focus and of double size.
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