For the two parallel rays \[AB\] and \[DE\] shown here, \[BD\] is the wave front. For what value of wavelength of rays destructive interference takes place between ray \[DE\] and reflected ray \[CD\]? |
A) \[\sqrt{3x}\]
B) \[\sqrt{2x}\]
C) \[x\]
D) \[2x\]
Correct Answer: A
Solution :
Path difference, |
\[\Delta x=(BC+CD)+\frac{\lambda }{2}\] |
Where \[CD=\frac{x}{\cos 30{}^\circ }=\frac{2x}{\sqrt{3}},\] |
And \[BC=CD\sin 30{}^\circ =\frac{2x}{\sqrt{3}}\times \frac{1}{2}=\frac{x}{\sqrt{3}}\] |
Now \[\Delta x=\left( \frac{x}{\sqrt{3}}+\frac{2x}{\sqrt{3}} \right)+\frac{\lambda }{2}=\sqrt{3}x+\frac{\lambda }{2}\] |
For destructive interference \[\Delta x=\frac{3\lambda }{2}\](here ) |
\[\therefore \sqrt{3}x+\frac{\lambda }{2}=\frac{3\lambda }{2}\] Or \[\lambda =\sqrt{3x}\] |
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