In a Young's double slit experiment, the slit separation d is 0.3 mm and the screen distance D is 1 m . A parallel beam of light of wavelength 600 nm is incident on the slits at angle α as shown in figure. On the screen, the point O is equidistant from the slits and distance PO is 11.0 mm . Which of the following statements(s) is/are correct?

For α=0.36/π degree, there will be destructive interference at point P .

For α=0.36/π degree, there will be destructive interference at point O .

For α=0 , there will be constructive interference at point P .

Physics · Wave Optics

JEE Advanced 2019 — Paper 2 — Question 1

In a Young's double slit experiment, the slit separation d is 0.3 mm and the screen distance D is 1 m . A parallel beam of light of wavelength 600 nm is incident on the slits at angle $\alpha$ as shown in figure. On the screen, the point O is equidistant from the slits and distance PO is 11.0 mm . Which of the following statements(s) is/are correct?

Option A:
For $\alpha=\frac{0.36}{\pi}$ degree, there will be destructive interference at point $O$ .
Option B:
Fringe spacing depends on $\alpha$ .
Option C:
For $\alpha=\frac{0.36}{\pi}$ degree, there will be destructive interference at point $P$ .
Correct
Option D:
For $\alpha=0$ , there will be constructive interference at point $P$ .

Answer: C

Step-by-step solution

Given $\alpha=\frac{0.36}{\pi}$ degree $=\frac{1}{500}$ radian.

Path difference at $\mathrm{O}=\mathrm{d} \sin \alpha=0.3 \times 10^{-3} \times \frac{1}{500}=6 \times 10^{-7}=\lambda$

Path difference at $\mathrm{P}=\mathrm{d}(\sin \alpha+\sin \theta)$

$=\mathrm{d} \frac{1}{500}+\frac{11}{1000}-$

$=\frac{\mathrm{d}}{500}-1+\frac{11}{2}-$

$=\lambda(6.5)$

For $\alpha=0$ , path difference at $P=5.5 \lambda$

$\therefore$ option C is correct

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2019
Paper: Paper 2
Subject: Physics
Chapter: Wave Optics
Topic: Young's Double Slit Experiment and Its Modifications