Question

What is the resulting intensity of two incoherent light bulbs each of intensity I ?

Answer 1

Hey mate!!

Here is your answer..

let us consider there are two needles say S1 and S2 moving up and down on the surface of the water and are pointing at point P. So the path difference here is given as S1P – S2P. Now the displacement by two needles and S1 S2 are:

y1 = A cos wt ……………… (1)

y2 = A cos wt …………….. (2)

So the resultant displacement at point P is, y = y1 + y2. When we substitute the value of y1 and y2 we write,

y = A cos wt + A cos wt

y = 2A cos wt……………….. (3)

Now, we know the intensity is proportional to the square of the amplitude waves.

I0 ∝ A²

Where I0 is the initial intensity and A² is the amplitude of the wave. From equation 3, we say that A = 2A. So,

I0 ∝ (2A)² or I0 ∝ 4 A²

I = 4 I0

Now, if two needles that are S1 are S2 are in the same phase, the potential difference is,

S1P – S2P = nλ

Where n = 0, 1, 2,3 ……… and λ = the wavelength of the wave. If the two needles S1 and S2 are vibrating at its destructive interference then, the potential difference is

S1P – S2P = (n + 1/2) λ

Now if the potential difference of the waves is Φ then,

y1 = α cos wt

y2 = α cos wt

The individual intensity of each wave is I0 , we get,

y = y1 + y2

= α cos wt + α cos (wt +Φ)

y = 2 α cos(Φ/2) cos (wt + Φ/2)

Since, the intensity is I0 ∝ A²

I0 ∝ 4α² cos² (Φ/2)

I = 4 I0 cos² (Φ/2)

Well, the time-averaged value of cos²(Φt/2) is 1/2. So, the resultant intensity will be I = 2 I0 at all the points.

Hope this helps you.

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