Two harmonic waves of monochromatic lighty_ 1 = a and y_ 2 = a ( … - Vidyadip

Question

Two harmonic waves of monochromatic light$\text{y}_{1} = \text{a} \cos\omega\text{t}\text{ and } \text{y}_{2} = \text{a} \cos(\omega\text{t} + \Phi)$
are superimposed on each other. Show that maximum intensity in interference pattern is four times the intensity due to each slit. Hence write the conditions for constructive and destructive interference in terms of the phase angle$\Phi$.

✓

Answer

Resultant displacement $\text{y} = \text{y}_{1} + \text{y}_{2}$ $ = \text{a}[ \cos(\omega\text{t}) + \cos(\omega\text{t} + \phi)]$ $ = 2 \text{ a}\cos\big(\frac{\phi}{2}\big)\cos\big(\omega\text{t} +\frac{\phi}{2}\big)$ $\therefore$ amplitude of resultant wave $= 2 \text{ a}\cos(\frac{\phi}{2} )$$\therefore$ Intensity $ = 4 \text{I}_{o}\cos^{2}(\frac{\phi}{2}) , \text{where } \text{I}_{o} =\text{a}^{2}$ is the intensity of each harmonic wave
At the maxima, $\phi =\pm2\text{n}\pi\therefore\cos^{2}\frac{\phi}{2} = 1 $ At the maxima, $\text{I} = 4 \text{ I}_{o} = 4 \times\text{intensity due to one slit }$ $\text{I} = 4 \text{ I}_{o}\cos^{2}(\frac{\phi}{2})$ For constructive interference, I is maximum It is possible when $\cos^{2}(\frac{\phi}{2}) =1 ; \frac{\phi}{2} =\text{n}\pi; \phi = 2 \text{n}\pi$ For destructive interference, I is minimum, i.e, I=0 It is possible when $\cos^{2}(\frac{\phi}{2}) = 0 ; \frac{\phi}{2} = \frac{(2\text{n} - 1 )\pi}{2}; \phi = (2\text{n}\pm1 )\frac{\phi}{2}$.

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