A) \[2f\]
B) \[\frac{3f}{2}\]
C) \[\frac{5f}{2}\]
D) \[3f\]
Correct Answer: B
Solution :
[b]| \[\frac{m=-\left( \frac{v}{u} \right)\Rightarrow \,\,\,-2=\frac{v}{u}}{2}\] |
| [\[-ve\] magnification because image is real] |
| \[\Rightarrow \] \[v=2u\] |
| From mirror formula, \[\left( \frac{1}{v} \right)+\left( \frac{1}{u} \right)=\frac{1}{f}\] \[\Rightarrow \] \[\left( \frac{1}{2u} \right)+\left( \frac{1}{u} \right)=\left( \frac{1}{f} \right)\] \[\Rightarrow \] \[\frac{3}{2u}=\frac{1}{f}\,\,\Rightarrow \,u=\frac{3f}{2}\] |
| [a] Only \[{{M}_{1}}\] is effective in making image since light rays will not fall on \[{{M}_{2}}\]. |
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