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) $
BETA
