question_answer

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 \[\lambda =\sqrt{3x}\]