shown three equidistant slits being illuminated by a monochromatic parallel beam of light. Let $$B{P_0} - A{P_0} = \lambda /3$$ and $$d > > \lambda $$ . show that in this case $$d = \sqrt {2\lambda D/3} $$ .(b) show that the intensity at $${P_0}$$ is three times the intensity due to any of the three slits individually.
Since, $$S_1. S_2$$ are in the same phase, at $$O$$ there will be maximum intensity.
Given that, there will be a maximum intensity at $$P$$.
The path difference between the two waves is:
$$BP_0-AP_0=\dfrac{\lambda}{3}$$
From the figure.
$$(\sqrt {D^2 +d^2})^2-D=\dfrac{\lambda}{3}$$
$$D^2+d^2=\dfrac{\lambda^2}{9}+D^2+\dfrac{2\lambda D}{3}$$
SInce $$(\lambda<
Above expression is simplified as:
$$d^2=\dfrac{2\lambda D}{3}$$
$$d=\sqrt{\dfrac{2\lambda D}{3}}$$
(b).
The path difference is given as $$BP_0-AP_0=\dfrac{\lambda}{3}=(\Delta x)_{12}$$
The phase difference is obtained as:
$$\phi_{12}=\dfrac{2\pi}{\lambda}\cdot\dfrac{\lambda}{3}=120^\circ$$
The path difference between the waves is :
$$CP_0-AP_0=\sqrt{(2d)^2+D^2}-D$$
$$\therefore \Delta x_{13}=\left[D\sqrt{1+\dfrac{2d}{D}}\right]^{1/2}$$
Simplify the abve equation by binomila expansion:
$$\Delta x_{13}=\dfrac{2d^2}{D}$$
We know that, $$d^2=\dfrac{2\lambda D}{3}$$, substitute the value in above equation.
$$\Delta x_{13}=2\times\left(\dfrac{2\lambda D}{3}\right)^2\times \dfrac 1D$$
$$\quad=\dfrac{4\lambda}{3}$$
So, the phase difference is
$$\phi_{13}=\dfrac{2\pi}{\lambda}\dfrac{4\lambda}{3}=480^\circ$$
$$\phi_{13}=480-2\pi=480-360=120^\circ$$
This shows that second and third wave are in same phase so their amplitude will be added up. so, the net amplitude of these two waves at phase difference of $$120^\circ$$ will be $$2A_0$$.
So net amplitude of three waves is:
$$A=\sqrt{A_1^2+A_2^2+2A_1A_2cos\theta}$$
$$A=\sqrt{A_0^2+4A_0^2+2A_0\times 2A_0cos120^\circ}=\sqrt3 A_0$$
Since intensity is given as:
$$I=A^2$$
i.e. $$I=(\sqrt3A_0)^2$$
So, the final intensity will be 3 times the initial intensity.
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