Question

"In a Young’s double slit interference experiment, the fringe pattern is observed on a screen placed at a distance D from the slits. The slits are separated by a distance d and are illuminated by monochromatic light of wavelength

λ. Find the distance from the central point where the intensity falls to (a) half the maximum, (b) one-fourth the maximum."

Answer 1

Given:

Separation between the two slits = d

Wavelength of the light = λ

Distance of the screen = D

(a) When the intensity is half the maximum:

Let Imax be the maximum intensity and I be the intensity at the required point at a distance y from the central point.

So,

I/Imax = 1/2

4a² cos ²[Ф/2] / 4 a² = 1/2

cos² [Ф/2] = 1/2

cos [ Ф/2] =1/√2

Ф/2 = π/4

Ф =π/2

Path difference x= λ/4

y= xD/ d= λD/ 4d

ii) When Intensity is 1/4 th maximum

I/ I max = 1/4

4a² cos ² (Ф/2) = 1/4

cos ²[Ф/2 )=1/4

cos (Ф/2) =1/2

Ф/2= π/3

Path difference = x=λ/3

y= xD/d= λD/3d

Answer 2

Answer:

thanks again few you sir roll of class 7