Question

In Young's double slit experiment using monochromatic light of wavelength λ, the maximum intensity of light at a point on the screen is K units. The intensity of light at point where the path difference is `lambda/3` is ______.

`[cos60^circ = sin30^circ  1/2]`

विकल्प

• `K/4`

• `K/2`

• K

• `(3K)/4`

Answer 1

In Young's double slit experiment using monochromatic light of wavelength λ, the maximum intensity of light at a point on the screen is K units. The intensity of light at the point where the path difference is `lambda/3` is `underlinebb(K/4)`.

Explanation:

For monochromatic light the resultant intensity is,

`l_R = l_1 + l_2 + 2sqrt(l_1l_2)costheta = 2l + 2l costheta` (∵ l₁ = l₂ = l) ..........(i)

For maximum intensity, θ = 0°

l_R = 2l + 2l cos0° = 4l (∵ cos0° = 1)

or K = 4l

or l = `K/4` .......(ii)

For path difference λ/3, phase difference,

`Phi = 2pi xx "Path difference"/lambda = 2pi xx (lambda"/"3)/lambda = (2pi)/3`

∴ `l_R^' = 2l + 2l cos  (2pi)/3` [From Eq. (i)]

= `2l + 2l(-1/2) = 2l - l = l`

= `K/4`           [using Eq. (ii)]