question_answer

A beam of light propagates through medium 1 and falls onto another medium 2, at an angle \[{{\alpha }_{1}}\] as shown in figure. After that, it propagates in Medium 2 at an angle \[{{\alpha }_{2}}\] as shown. The light's wavelength in medium 1 is \[{{\lambda }_{1}}\]. What is the wavelength of light in medium 2?

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] \[{{\mu }_{1}}\sin \alpha ={{\mu }_{2}}\sin {{\alpha }_{2}}\] \[\frac{c}{{{v}_{1}}}\sin {{\alpha }_{1}}=\frac{c}{{{v}_{2}}}\sin {{\alpha }_{2}}\] \[\frac{\sin {{\alpha }_{1}}}{f{{\lambda }_{1}}}=\frac{\sin {{\alpha }_{2}}}{f{{\lambda }_{2}}}\] \[{{\lambda }_{2}}=\frac{\sin {{\alpha }_{2}}}{\sin {{\alpha }_{1}}}{{\lambda }_{1}}\]