Optics Question 898

Question: A light wave of wavelength $ {\lambda _{0}} $ propagates from point A to point B. We introduce in its path a glass plate of refractive index n and thickness l. The introduction of the plate alters the phase of the plate at B by an angle $ \phi $ . If $ \lambda $ is the wavelength of lights on emerging from the plate, then

Options:

A) $ \Delta \phi =0 $

B) $ \Delta \phi =\frac{2\pi l}{{\lambda _{0}}} $

C) $ \Delta \phi =2\pi {{\ln }^{2}}( \frac{1}{\lambda }-\frac{1}{{\lambda _{0}}} ) $

D) $ \Delta \phi =\frac{2\pi l}{{\lambda _{0}}}(n-1) $

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Answer:

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 ( \frac{1}{\lambda }-\frac{1}{{\lambda _{0}}} ) $

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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