Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Wave Optics

Question:

In the fig. shown CP represents the plane wave front and AO and BP, the corresponding two rays. Find the condition on θ for constructive interference at P between the ray BP and reflected ray OP

Options:

$\cos \theta=\frac{3 \lambda}{2 d}$

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

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

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

Correct Answer:

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

Explanation:

$PN = d, P O=d \sec \theta$

$CO = PO \cos 2 \theta=d \sec \theta \cos 2 \theta$

Path difference = CO + PO = $d \sec \theta+d \sec \theta \cos 2 \theta$

$=d \sec \theta(1+\cos 2 \theta)=\frac{d}{\cos \theta} \times 2 \cos ^2 \theta=2 d \cos \theta$

The ray AO is incident on denser medium, so reflected ray suffers a phase difference of $\pi$ or a path differences of $\frac{\lambda}{2}$

∴ Effective path difference $\Delta=2 d \cos \theta+\frac{\lambda}{2}$

For constructive interference path difference $\Delta=n \lambda$.

$\Rightarrow 2 d \cos \theta+\frac{\lambda}{2}=n \lambda$

or $2 d \cos \theta=(2 n-1) \frac{\lambda}{2}$