$P Q$ represents a wavefront and $A O$ and $B P$, the corresponding two rays. Find the condition on $Q$ for constructive interference at $P$ between ray $B P$ and reflected ray $O P$.

Question

$P Q$ represents a wavefront and $A O$ and $B P$, the corresponding two rays. Find the condition on $Q$ for constructive interference at $P$ between ray $B P$ and reflected ray $O P$., $\cos \theta=\frac{3 \lambda}{2 d}$, $\cos \theta=\frac{\lambda}{4 d}$, $\sec \theta-\cos \theta=\frac{\lambda}{d}$, $\sec \theta-\cos \theta=\frac{4 \lambda}{d}$

MARKS App · Accepted Answer

According to figure, point $P$ and point $Q$ are at same phase. According to figure, In $\triangle P O R$, $\begin{aligned} & \cos \theta=\frac{P R}{O P}=\frac{d}{O P} \ & \Rightarrow \quad O P=\frac{d}{\cos \theta} \quad \ldots (i) \end{aligned}$ In $\triangle Q O P$, $\begin{aligned} & & \sin \left(90^{\circ}-2 \theta\right) & =\frac{O Q}{O P} \ \Rightarrow & & \cos 2 \theta & =\frac{O Q}{O P} \ \Rightarrow & & O Q & =O P \cos 2 \theta \quad \ldots (ii) \end{aligned}$ $\therefore$ Path difference, $\begin{aligned} & \Delta=O P+O Q=O P+O P \cos 2 \theta \ & =O P(1+\cos 2 \theta) \quad\left[\because 1+\cos 2 \theta=2 \cos ^2 \theta\right] \ & =\frac{d}{\cos \theta} \cdot 2 \cos ^2 \theta \quad \text { [from Eq. (i)] } \ & =2 d \cos \theta \end{aligned}$ But path difference is $\frac{\lambda}{2}$. $\begin{aligned} & \therefore & 2 d \cos \theta & =\frac{\lambda}{2} \ \Rightarrow & & \cos \theta & =\frac{\lambda}{4 d} \end{aligned}$

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