Question

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

• A. $\cos \theta =\frac{3\lambda }{2d}$
• B. $\cos \theta =\frac{\lambda }{4d}$
• 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 $△POR$
$\cos \theta =\frac{PR}{OP}=\frac{d}{OP}$
$\Rightarrow OP=\frac{d}{\cos \theta }...(i)$
In $\Delta QOP$
$\sin \left(90^{\circ }-2\theta \right)=\frac{OQ}{OP}$
$\Rightarrow \cos 2\theta =\frac{OQ}{OP}$
$\Rightarrow OQ=OP\cos 2\theta \dots$ (ii)
$\therefore$ Path difference,
$\Delta =OP+OQ=OP+OP\cos 2\theta$
$=OP(1+\cos 2\theta )\left[\because 1+\cos 2\theta =2{\cos }^2\theta \right]$
$=\frac{d}{\cos \theta }\cdot 2{\cos }^2\theta [$ from $Eq$. (i) $]$
$=2d\cos \theta$
But path difference is $\frac{\lambda }{2}$.
$\therefore 2d\cos \theta =\frac{\lambda }{2}$
$\Rightarrow \cos \theta =\frac{\lambda }{4d}$