A) \[\sqrt{2}\frac{\sin 30{}^\circ }{\sin 60{}^\circ }\]
B) \[\sqrt{2}\frac{\sin 45{}^\circ }{\sin 60{}^\circ }\]
C) \[\sqrt{2}\frac{\sin 15{}^\circ }{\sin 60{}^\circ }\]
D) \[\sqrt{2}\frac{\sin 60{}^\circ }{\sin 30{}^\circ }\]
Correct Answer: B
Solution :
[c] Snell's law at C \[1\times \sin 90{}^\circ =\sqrt{2}\,\sin {{r}_{2}}\] \[{{r}_{2}}=45{}^\circ \] Snell's law at B \[\sqrt{2}\sin 15{}^\circ =1\sin r\] ...(i) Snell's law at A \[\mu .\sin 60{}^\circ =1.\sin r\] ...(ii) From (i) and (ii), \[\mu .\sin 60{}^\circ =\sqrt{2}\sin 15{}^\circ \] \[\mu =\sqrt{2}\frac{\sin 15{}^\circ }{\sin 60{}^\circ }\]You need to login to perform this action.
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