Question

Amplitude of waves observed by two light sources of same wave length are \( a \) and \( 2 a \) and have a phase difference of \( \pi \) between them. Then minimum intensity of light will be proportional to 

a) 0

b) \( 5 a^{2} \)

c) \( a^{2} \)

d) \( 9 a^{2} \)

Answer 1

Correct option is c) a²

According to the question, it is given that,

Amplitude of first source, A₁ = a

Amplitude of second source, A₂ = 2a

And phase difference, \(\phi\) = π

We know that, intensity ∝ (Amplitude)².

i.e, I ∝ A ⇒ I = KA²

So, I₁ = Ka²

And I₂ = 4Ka²

Now for minimum intensity, 

I_min = \(\sqrt{I_1} - \sqrt{I_2}^2\)

\(=\sqrt{Ka^2} - \sqrt{4Ka^2}^2\)

\(= a\sqrt K - 2aK^2\)

\(= -a\sqrt K^2\)

\(= Ka^2\)

Hence, minimum intensity is directly proportional to a².