A) \[\frac{2\pi }{\lambda }\left( \frac{{{L}_{1}}}{{{\mu }_{1}}}-\frac{{{L}_{2}}}{{{\mu }_{2}}} \right)\]
B) \[\frac{2\pi }{\lambda }\left( {{L}_{2}}-{{L}_{1}} \right)\]
C) \[\frac{2\pi }{\lambda }\left( {{\mu }_{2}}{{L}_{1}}-{{\mu }_{1}}{{L}_{2}} \right)\]
D) \[\frac{2\pi }{\lambda }\left( {{\mu }_{1}}{{L}_{1}}-{{\mu }_{2}}{{L}_{2}} \right)\]
Correct Answer: D
Solution :
Optical path for first ray = \[{{\text{R}}_{\text{1}}}\text{:}{{\text{R}}_{\text{2}}}\text{:}{{\text{R}}_{\text{3}}}\text{=}\frac{{{\text{I}}_{\text{1}}}}{{{\text{A}}_{\text{1}}}}\text{:}\frac{{{\text{I}}_{\text{2}}}}{{{\text{A}}_{\text{2}}}}\text{:}\frac{{{\text{I}}_{\text{3}}}}{{{\text{A}}_{\text{3}}}}\] Optical path for second ray = \[\text{=}\frac{\text{I}_{\text{1}}^{\text{2}}}{{{\text{V}}_{\text{1}}}}\text{:}\frac{\text{I}_{\text{2}}^{\text{2}}}{{{\text{V}}_{\text{2}}}}\text{:}\frac{\text{I}_{\text{3}}^{\text{2}}}{\left( {{\text{m}}_{\text{3}}}\text{d} \right)}\] So, phase difference is given by \[=\frac{I_{1}^{2}}{\left( {{m}_{1}}d \right)}:\frac{I_{2}^{2}}{\left( {{m}_{2}}d \right)}:\frac{I_{3}^{2}}{\left( {{m}_{3}}d \right)}\] \[=\frac{I_{1}^{2}}{{{m}_{1}}}:\frac{I_{2}^{2}}{{{m}_{2}}}:\frac{I_{3}^{2}}{{{m}_{3}}}\] \[=\frac{{{5}^{2}}}{2}:\frac{{{3}^{2}}}{3} & :\frac{{{2}^{2}}}{5}\] \[=\frac{25}{2}:3:\frac{4}{5}=125:30:8\]
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