question_answer

                A small coin is resting on the bottom of a beaker filled with a liquid. A ray of light from the coin travels upto the surface of the liquid and moves along its surface (see figure).                                            

A)                                 How fast is the light travelling in the liquid?                 \[1.8\times {{10}^{8}}\,m/s\]

B)                 \[2.4\times {{10}^{8}}\]

C)                 \[3.0\times {{10}^{8}}\,m/s\]

D)                 \[1.2\times {{10}^{8}}\,m/s\]

Correct Answer: A

Solution :

                Key Idea: Critical angle is the angle of incidence in denser medium for which the angle of refraction in rarer medium is \[{{90}^{\text{o}}}\].                 As shown in figure, a light ray from the coin will not emerge out of liquid, if \[i>C\].                                 Therefore, minimum radius R corresponds to i = C. In \[\Delta \,\,SAB,\]                 \[\frac{R}{h}=\tan C\] or            R = h tan C \[orR=\frac{h}{\sqrt{{{\mu }^{2}}-1}}\]                 Given,   R = 3 cm, h = 4 cm Hence,                  \[\frac{3}{4}=\frac{1}{\sqrt{{{\mu }^{2}}-1}}\] or            \[{{\mu }^{2}}=\frac{25}{9}\,or\,\mu =\frac{5}{3}\] But         \[\mu =\frac{c}{v}or\,v=\frac{c}{\mu }\]                 \[=\frac{3\times {{10}^{8}}}{5/3}\]                 \[=1.8\times {{10}^{8}}\,m/s\]