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

In a Young's double slit experiment, the intensity at some point on the screen is found to be \(\dfrac{3}{4}\) times of the maximum of the interference pattern. The path difference between the interfering waves at this point is \(\dfrac{\lambda}{x}\) where \(\lambda\) is wavelength of the incident light. The value of \(x\) is _______.

  1. A 1
  2. B 3
  3. C 5
  4. D 6
Verified Solution

Answer & Solution

Correct Answer

(D) 6

Step-by-step Solution

Detailed explanation

The intensity at a point in Young's double slit experiment is given by \(I = I_{max} \cos^2\left(\dfrac{\phi}{2}\right)\) Given \(I = \dfrac{3}{4} I_{max}\) \(\dfrac{3}{4} I_{max} = I_{max} \cos^2\left(\dfrac{\phi}{2}\right)\)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app