A) \[\sqrt{3}\times {{10}^{8}}\,\,m/s\]
B) \[6\times {{10}^{8}}\,\,m/s\]
C) \[1.5\times {{10}^{8}}\,\,m/s\]
D) \[3\times {{10}^{8}}\,\,m/s\]
Correct Answer: C
Solution :
Key Idea: Angle of incidence becomes critical angle when angle of refraction is\[{{90}^{\circ }}\]. When a ray of light passes from an optically denser to rarer medium, then the angle of refraction \[r\] is greater than corresponding angle of incidence \[i\] and critical angle is obtained when \[r={{90}^{\circ }}\] \[\frac{\sin \,{{\theta }_{c}}}{\sin \,{{90}^{\circ }}}=\frac{1}{\mu }\] \[\Rightarrow \] \[\mu =\frac{1}{\sin \,\,{{\theta }_{c}}}\] Given, \[{{\theta }_{c}}={{30}^{\circ }}\] \[\therefore \] \[\mu =\frac{1}{\sin \,\,{{30}^{\circ }}}=2\] From Snell?s law \[\mu =\frac{{{v}_{v}}}{{{v}_{m}}}\] \[\Rightarrow \] \[{{v}_{m}}=\frac{{{v}_{v}}}{\mu }=\frac{3\times {{10}^{8}}}{2}\] \[=1.5\times {{10}^{8}}\,\,m/s\] Note: Velocity is lesser in the denser medium
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