In a Young's double slit experiment, the intensity at some point on the screen is found to be 34 times of the maximum of the interference pattern. The path difference between the interfering waves at this point is x where is wavelength of the incident light. The value of x is .
Answer
6
Solution
I = Imax 2 2 = 34Imax
⇒ 2 = 32 ⇒ 2 = 6 ⇒ = 3
Path difference x = 2 = 2 3 = 6
∴ x = 6
Answer: 6