The equation of two interfering light waves are
$y_{1} = b cos ω t$ and $y_{2} = b sin (ω t + Φ)$ .
If the light waves interfere destructively, then the possible value of $Φ$ are

Question

The equation of two interfering light waves are y 1 ​ = b cos ω t and y 2 ​ = b sin ( ω t + Φ ) . If the light waves interfere destructively, then the possible value of Φ are

TestHub · Accepted Answer

The given equations of the light waves are:

$y_{1} = b cos ω t = b sin (ω t + \frac{π}{2})$

$y_{2} = b sin (ω t + Φ)$

The phase difference between the two waves is:

$Δ ϕ = (ω t + Φ) - (ω t + \frac{π}{2}) = Φ - \frac{π}{2}$

For destructive interference, the phase difference must be an odd multiple of $π$ :

$Δ ϕ = (2 n - 1) π$

$Φ - \frac{π}{2} = (2 n - 1) π$

$Φ = (2 n - 1) π + \frac{π}{2} = \frac{( 4 n - 1 ) π}{2}$

Substituting $n = 1, 2, 3, \dots$ , we get the possible values of $Φ$ :

$Φ = \frac{3 π}{2}, \frac{7 π}{2}, \frac{11 π}{2}, \dots$

Answer: $\frac{3 π}{2}, \frac{7 π}{2}, \frac{11 π}{2}, \dots\dots$

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