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  • In the below diagram, CP represents a wavefront and AO and BP, the corresponding two rays. Find the condition of

In the below diagram, CP represents a wavefront and AO and BP, the corresponding two rays. Find the condition of \mathrm{\theta} for constructive interference at P between the rays BP and reflected ray AOP:

Option: 1

\mathrm{\cos \theta=3 \lambda / 2 d}


Option: 2

\mathrm{\cos \theta=\lambda / 4 d}


Option: 3

\mathrm{\sec \theta-\cos \theta=\lambda / \mathrm{d}}


Option: 4

\mathrm{\sec \theta-\cos \theta=4 \lambda / d}


Answers (1)

best_answer

Path difference between the two rays is given by

\mathrm{ \begin{array}{ll} & \Delta=\mathrm{CO}+\mathrm{PO} \\\\ \because & \mathrm{PR}=\mathrm{d}, \\\\ \text { So } & \mathrm{PO}=\mathrm{d} \sec \theta \\\\ \text { and } & \mathrm{CO}=\mathrm{PO} \cos 2 \theta=\mathrm{d} \sec \theta \cos 2 \theta \\\\ \text { So, } & \Delta=(\mathrm{d} \sec \theta+\mathrm{d} \sec \theta \cos 2 \theta) \end{array} }

Phase difference between two rays is \phi=\pi (as one ray is reflected one and another is direct). Now, for constructive interference, path difference should be even multiple of half wavelength.

i.e. \mathrm{\Delta=\lambda / 2,}            \mathrm{3 \lambda / 2, \ldots}
So, \mathrm{\quad \mathrm{d} \sec \theta+\mathrm{d} \sec \theta \cos 2 \theta=\frac{\lambda}{2}}

\mathrm{\text { or } \quad \mathrm{d} \sec \theta(1+\cos 2 \theta)=\frac{\lambda}{2}}

\mathrm{\text { or } \quad \mathrm{d} \sec \theta\left(2 \cos ^2 \theta\right)=\frac{\lambda}{2}}

\mathrm{\therefore \quad \cos \theta=\frac{\lambda}{4 d}}

 

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Gaurav

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