Answer:
Given: \[m=-2,\,v=-30\,cm\]
We know that, \[m=\frac{-v}{u}\]
\[-2=\frac{30}{u}\]
\[u=-15cm\]
Now, \[f=\frac{uv}{u+v}=\frac{-15\,\times -30}{-15-30}=\frac{450}{-45}=-10cm\]
If the object is shifted 10 cm towards the mirror then, \[u=-5\,cm,\] i.e., object is between pole and focus, thus image formed will be virtual, erect and magnified.
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