A) at one of the foci, virtual and double its size
B) at\[\frac{3f}{2}\], real and inverted
C) at\[2f\], virtual and erect
D) none of the above
Correct Answer: A
Solution :
From lens formula \[\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\] (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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