A) 12 cm, 0.2 cm
B) 18 cm, 0.2 cm
C) 6 cm, 0.5 cm
D) 5 cm, 0.1 cm
Correct Answer: A
Solution :
[a]Here, Object distance, \[u=-\text{ }30cm\] |
Focal length, \[f=+20\text{ }cm\] |
We know, mirror formula, \[\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\] \[\Rightarrow \] \[v=\frac{uf}{u-f}=\frac{-30\times 20}{-30-20}=+12\,cm\] |
Again, magnification, \[m=\frac{-v}{u-f}=\frac{-12\,cm}{30\,cm}=\frac{2}{5}\] |
Now, Image height \[=m\times \] object height \[=\frac{2}{5}\times 0.5\,cm=0.2\,cm\] |
Thus the image is formed behind the mirror at a distance of 12 cm from the pole. Image height is 0.2 cm. |
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