In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. What is the intensity of light at a point where path difference is λ/3?

Let I₁and I₂ be the intensity of the two lightwaves. Their resultant intensities can be obtained as: I’ = I₁ + I₂ + 2√I₁l₂  cos 0

Where, 0 = Phase difference between the two waves For monochromatic light waves,

I₁ = I₂

Therefore , I’ = I₁ + I₁+ 2 √I_1I1  cos 0

= 2I₁+2I₁ cos0

Phase difference = 2π/λ   x path difference

Since path difference = λ,

Phase difference, 0 = 2π

Therefore ,I’ =2I₁+2I₁ =4 I₁

Given I’ = K

Therefore ,I₁= K/4 ——-Eq-1
When path difference = λ/3, the phase difference, 0 =2π/3

Hence, resultant intensity, I’_R = I₁ + I₂ + 2√I₁I₂ cos 2π/3

= 2I₁ + 2I₁ (-1/2) = I₁

Using equation (1), we can write:

I_R = I₁=K/4

Hence, the intensity of light at a point where the path difference is λ/3 is K/4 units.