Two identical radiators have a separation of d=λ / 4 where λ is the wavelength of the waves emitted by either source. The initial phase difference between the source is π / 4. Then the intensity on the screen at a distant point situated at an angle, θ=30° from the radiators is (here, I0 is intensity at that point due to one radiator alone)

Question

Two identical radiators have a separation of d=λ / 4 where λ is the wavelength of the waves emitted by either source. The initial phase difference between the source is π / 4. Then the intensity on the screen at a distant point situated at an angle, θ=30° from the radiators is (here, I0 is intensity at that point due to one radiator alone) (A) I0 (B) 2 I0 (C) 3 I0 (D) 4 I0

Tardigrade Admin · Accepted Answer

The intensity at a point on screen is given by

I = 4 I_{0} cos^{2} (ϕ /2)

When

ϕ

is the phase difference.
In this problem

ϕ

arises (i) due to initial phase difference of

\frac{π}{4}

and (ii) due to path difference for the observation point situated at

θ = 3 0^{\circ}

.
Thus,

ϕ = \frac{π}{4} + \frac{2 π}{λ} (d sin θ)

= \frac{π}{4} + \frac{2 π}{λ} - \frac{λ}{4} (sin 3 0^{\circ})

= \frac{π}{4} + \frac{π}{4} = \frac{π}{2}

Hence,

\frac{ϕ}{2} = \frac{π}{4}

and

I = 4 I_{0} cos^{2} (\frac{π}{4}) = 2 I_{0}

$I_{0}$

$2 I_{0}$

$3 I_{0}$

$4 I_{0}$

I0​

2I0​

3I0​

4I0​

$I_{0}$

$2 I_{0}$

$3 I_{0}$

$4 I_{0}$