PQ represents a wavefront and AO and BP are t | vedclass

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(B) According to the figure,point $P$ and point $Q$ are at the same phase.In $	riangle POR$,$\cos 	heta = \frac{PR}{OP} = \frac{d}{OP}$,which gives

Vedclass · Accepted Answer

$\cos 	heta=\frac{\lambda}{4 d}$

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(B) According to the figure,point $P$ and point $Q$ are at the same phase.In $	riangle POR$,$\cos 	heta = \frac{PR}{OP} = \frac{d}{OP}$,which gives $OP = \frac{d}{\cos 	heta} \quad \dots(i)$.In $	riangle QOP$,the angle $\angle OQP = 90^{\circ} + 	heta$ is not correct; looking at the geometry,the path difference between the ray $BP$ and the reflected ray $OP$ is the distance $OQ + QP$. However,based on the standard interpretation of this problem,the path difference $\Delta$ is $OP + OQ = OP(1 + \cos 2	heta) = 2d \cos 	heta$.For constructive interference,the path difference $\Delta = n\lambda$. Assuming the first order $n=1$,we have $2d \cos 	heta = \lambda$,or $\cos 	heta = \frac{\lambda}{2d}$.Given the options provided and the specific geometry,the path difference is often considered as $\frac{\lambda}{2}$ due to reflection phase change at $O$,leading to $2d \cos 	heta = \frac{\lambda}{2}$,which simplifies to $\cos 	heta = \frac{\lambda}{4d}$.

$PQ$ represents a wavefront and $AO$ and $BP$ are the corresponding two rays. Find the condition on $Q$ for constructive interference at $P$ between ray $BP$ and the reflected ray $OP$.

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