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

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Let the amplitudes of the two identical light waves be $A_{1} = A_{2} = A$ .

The resultant amplitude $A_{R}$ after superposition is given by

$A_{R}^{2} = A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} cos ϕ$

Substituting $A_{1} = A_{2} = A$ , we get

$A_{R}^{2} = A^{2} + A^{2} + 2 A^{2} cos ϕ$

$A_{R}^{2} = 2 A^{2} (1 + cos ϕ)$

Using the trigonometric identity $1 + cos ϕ = 2 cos^{2} (\frac{ϕ}{2})$ , we obtain

$A_{R}^{2} = 4 A^{2} cos^{2} (\frac{ϕ}{2})$

Since the intensity $I$ of a wave is directly proportional to the square of its amplitude ( $I \propto A_{R}^{2}$ ), the resultant intensity is

$I_{R} \propto cos^{2} (\frac{ϕ}{2})$

Answer: $cos^{2} (\frac{ϕ}{2})$

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