question_answer

                  Refractive index of water is 5/3. A light source is placed in water at a depth of 4 m. Then what must be the minimum radius of disc placed on water surface so that the light of source can be stopped?

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 \[\Delta SAB\]                 \[\frac{R}{h}=\tan {{\theta }_{C}}\]  \[\therefore R=h\tan {{\theta }_{c}}\]                                         \[orR=\frac{h}{\sqrt{{{\mu }^{2}}-1}}=\frac{4}{\sqrt{{{\left( \frac{5}{3} \right)}^{2}}-1}}\] \[=\frac{4\times 3}{\sqrt{25-9}}=\frac{4\times 3}{4}=3\,m\]