Question

Two interfering sources in YDSE have intensities 5I and 4I . The intensity at a point where the two lights have a path difference lamda by 4 is?
4I
5I
3I
none
​

Answer 1

Net intensity after interference of waves with phase difference ϕ

Inet = I 1 +I 2+2√ I 1 I 2 cosϕ

Thus I net 1 (ϕ=π/2)=5I

I net 2 (ϕ=π)=I

Thus, the difference between them =4I

Answer 2

Given:

Two interfering lights with intensities 5I and 4I

To Find:

Intensity at the point where path difference is λ/4

Solution:

Let Δx represent the path difference and ΔФ represents the phase difference between the given light intensities.

We know that in Young's Double Slit Experiment, the resultant intensity I due to interference of two intensities I₁ and I₂ at a point with phase difference ΔФ is given by:

I = $\ I_{1} + I_{2} + 2 \sqrt{I_{1}I_{2}} cos \phi$     - (1)

Also, Δx/λ = ΔФ/2π

According to the question,

$\frac{\lambda}{4}$ / λ = ΔФ/2π

or 2π/4 = ΔФ

or ΔФ = π/2

Substituting the value of ΔФ in (1),

I = 5I+4I+2( √20I) cos π/2

Since cos π/2 = cos 90° = 0

so I = 5I + 4I + 0

= 9I

Hence, the intensity at the given point is 9I i.e. none of the given options.