Question

$P Q$ represents a wave front 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. $\cos \theta=\frac{3 \lambda}{2 d}$
• B. $\cos \theta=\frac{\lambda}{4 d}$
• C. $\sec \theta-\cos \theta=\frac{\lambda}{d}$
• D. $\sec \theta-\cos \theta=\frac{4 \lambda}{d}$

Answer 1

Answer: (B)

According to figure, point $P$ and point $Q$ are at same phase.
According to figure,

In $\triangle P O R$
$\cos \theta=\frac{P R}{O P}=\frac{d}{O P}$
$\Rightarrow O P=\frac{d}{\cos \theta} ...(i)$
In $\Delta Q O P$
$\sin \left(90^{\circ}-2 \theta\right)=\frac{O Q}{O P}$
$\Rightarrow \cos 2 \theta=\frac{O Q}{O P}$
$\Rightarrow O Q=O P \cos 2 \theta \ldots$ (ii)
$\therefore $ Path difference,
$\Delta=O P+O Q=O P+O P \cos 2 \theta$
$=O P(1+\cos 2 \theta)\left[\because 1+\cos 2 \theta=2 \cos ^{2} \theta\right]$
$=\frac{d}{\cos \theta} \cdot 2 \cos ^{2} \theta [$ from $Eq$. (i) $]$
$=2 d \cos \theta$
But path difference is $\frac{\lambda}{2}$.
$\therefore 2 d \cos \theta=\frac{\lambda}{2}$
$\Rightarrow \cos \theta=\frac{\lambda}{4 d}$