A) 3 m
B) 4 m
C) 5 m
D) \[\infty \]
Correct Answer: A
Solution :
Key Idea: The light from the source will not emerge out of water if angle of incidence is greater than critical angle. As shown in figure, \[i>{{\theta }_{C}}\]. Therefore, minimum radius R corresponds to \[i>{{\theta }_{C}}\]. In \[\Delta SAB\] \[\frac{R}{h}=\tan {{\theta }_{C}}\] \[\therefore R=h\tan {{\theta }_{c}}\] \[orR=\frac{h}{\sqrt{{{\mu }^{2}}-1}}=\frac{4}{\sqrt{{{\left( \frac{5}{3} \right)}^{2}}-1}}\] \[=\frac{4\times 3}{\sqrt{25-9}}=\frac{4\times 3}{4}=3\,m\]You need to login to perform this action.
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