\(P Q\) 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 \(O P\).

• A

  \(\cos \theta=\frac{3 \lambda}{2 d}\)
• Correct answer
  B

  \(\cos \theta=\frac{\lambda}{4 d}\)
• C

  \(\sec \theta-\cos \theta=\frac{\lambda}{d}\)
• D

  \(\sec \theta-\cos \theta=\frac{4 \lambda}{d}\)

Solution

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 \(\frac{\lambda}{2}\). \(\begin{aligned} & \therefore & 2 d \cos \theta & =\frac{\lambda}{2} \\ \Rightarrow & & \cos \theta & =\frac{\lambda}{4 d} \end{aligned}\)

Hence, the correct answer is option: 'B'.