Two identical light waves having phase differenceϕ propagate in the same direction. When they superpose, the intensity of the resultant wave is proportional to

Question

Two identical light waves having phase differenceϕ propagate in the same direction. When they superpose, the intensity of the resultant wave is proportional to (A)  cos 2 φ (B)  cos 2 (φ/2) (C)  cos 2 (φ/3) (D)  cos 2 (φ/4)

Tardigrade Admin · Accepted Answer

The resultant intensity after the interference of the two light waves is

I_{R} = I_{1} + I_{2} + 2 I_{1} I_{2} cos ϕ

For two identical light waves,

I = I_{1} = I_{2}

Therefore,

I_{R} = I + I + 2 I^{2} cos ϕ = 2 I 1 + cos ϕ

\Rightarrow I_{R} = 4 I cos^{2} \frac{ϕ}{2}

∴ I_{R} \propto cos^{2} \frac{ϕ}{2}

Q. Two identical light waves having phase differenceϕ propagate in the same direction. When they superpose, the intensity of the resultant wave is proportional to

$cos^{2} ϕ$

$cos^{2} \frac{ϕ}{2}$

$cos^{2} \frac{ϕ}{3}$

$cos^{2} \frac{ϕ}{4}$

Q. Two identical light waves having phase differenceϕ propagate in the same direction. When they superpose, the intensity of the resultant wave is proportional to

cos2ϕ

cos22ϕ​

cos23ϕ​

cos24ϕ​

The resultant intensity after the interference of the two light waves is IR​=I1​+I2​+2I1​I2​​cosϕ For two identical light waves, I=I1​=I2​ Therefore, IR​=I+I+2I2​cosϕ=2I1+cosϕ ⇒IR​=4Icos22ϕ​ ∴IR​∝cos22ϕ​

$cos^{2} ϕ$

$cos^{2} \frac{ϕ}{2}$

$cos^{2} \frac{ϕ}{3}$

$cos^{2} \frac{ϕ}{4}$