A) \[\Delta \phi =0\]
B) \[\Delta \phi =\frac{2\pi l}{{{\lambda }_{0}}}\]
C) \[\Delta \phi =2\pi {{\ln }^{2}}\left( \frac{1}{\lambda }-\frac{1}{{{\lambda }_{0}}} \right)\]
D) \[\Delta \phi =\frac{2\pi l}{{{\lambda }_{0}}}(n-1)\]
Correct Answer: D
Solution :
[d] \[{{\phi }_{i}}=\frac{2\pi }{{{\lambda }_{0}}}\ell \] \[{{\phi }_{f}}=\frac{2\pi }{\lambda }\ell \] \[\Delta \phi =2\pi \ell \left( \frac{1}{\lambda }-\frac{1}{{{\lambda }_{0}}} \right)\] Further, by Snell's law \[n\lambda =(1){{\lambda }_{0}}\Rightarrow \lambda =\frac{{{\lambda }_{0}}}{n}\Rightarrow \Delta \phi =\frac{2\pi l}{{{\lambda }_{0}}}(n-1)\]
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