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Two light waves are represented by y₁ = a sinwt and y₂ =a sin(wt +d). The phase of the resultant wave is

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$\frac{δ}{2}$

$\frac{δ}{3}$

$\frac{δ}{4}$

answer is B.

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Detailed Solution

$I = 4 I_{0} c o s^{2} (\frac{δ}{2})$

$ϕ = \frac{δ}{2}$

$tanϕ = \frac{sinθ}{1 + cosθ} when two waves have same amplitude$

$on simplyfing tanϕ = \tan \frac{θ}{2}$

so $h e r e θ = δ s o ϕ = δ / 2$

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