ExamBro
ExamBro
JEE Mains · Physics · STD 12 - 10. Wave optics

In a Young’s double slit experiment, the path difference, at a certain point on the screen, between two interfering waves is \(1/8^{th}\) of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to

  1. A \(0.74\)
  2. B \(0.85\)
  3. C \(0.94\)
  4. D \(0.80\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(0.85\)

Step-by-step Solution

Detailed explanation

\(\Delta x=\frac{\lambda}{8}\) Phase \(|\Delta P|=\frac{2 \pi}{\lambda} \times \frac{\lambda}{8}=\frac{\pi}{4}\) \(\therefore \) \(\mathrm{I}_{\mathrm{res}}=I+I+2I \cos \left(\frac{\mathrm{R}}{4}\right)\) \(=\,2I\left(1+\frac{1}{\sqrt{2}}\right)=2I \times 1.7\)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app