A) \[42\,m\]
B) \[6\,m\]
C) \[2\sqrt{7m}\,\]
D) \[12m\]
Correct Answer: B
Solution :
If the bulb is not visible, it must be ray (light) coming from the bulb makes an angle equal or greater than critical angle, to the normal at the surface. So, \[\sin \theta \,\,c=\frac{1}{\mu }=\frac{r}{\sqrt{{{r}^{2}}+28}}\] \[\mu =\frac{4}{3}\] \[\Rightarrow \] \[r=6\,m\]You need to login to perform this action.
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