A) \[\frac{\sin \,{{\alpha }_{1}}}{\sin {{\alpha }_{2}}}{{\lambda }_{1}}\]
B) \[\frac{\sin \,{{\alpha }_{2}}}{\sin {{\alpha }_{1}}}{{\lambda }_{1}}\]
C) \[\frac{\cos \,{{\alpha }_{1}}}{\cos {{\alpha }_{2}}}{{\lambda }_{1}}\]
D) \[\frac{\cos \,{{\alpha }_{2}}}{\cos {{\alpha }_{1}}}{{\lambda }_{1}}\]
Correct Answer: B
Solution :
[b] \[{{\mu }_{1}}\sin \alpha ={{\mu }_{2}}\sin {{\alpha }_{2}}\] \[\frac{c}{{{v}_{1}}}\sin {{\alpha }_{1}}=\frac{c}{{{v}_{2}}}\sin {{\alpha }_{2}}\] \[\frac{\sin {{\alpha }_{1}}}{f{{\lambda }_{1}}}=\frac{\sin {{\alpha }_{2}}}{f{{\lambda }_{2}}}\] \[{{\lambda }_{2}}=\frac{\sin {{\alpha }_{2}}}{\sin {{\alpha }_{1}}}{{\lambda }_{1}}\]You need to login to perform this action.
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