Physics - Wave Optics Question with Solution | TestHub

Phase difference corresponding to path difference$\lambda$is

$\mathrm{\Delta }\varphi =\frac{2\pi }{\lambda }\times \lambda =2\pi$

Intensity at the point having this phase difference$2\pi$is

$I=4I_0{\cos }^2\frac{\varphi }{2}$

$=4I_0{\left(\cos \frac{2\pi }{2}\right)}^2$

$=4I_0{\left(-1\right)}^2=4I_0$

Here,$I_0$is intensity of individual source.

It is given that

$I=4I_0=K$

$\Rightarrow I_0=\frac{K}{4}$......(i)

Phase difference corresponding to path difference of$\frac{\lambda }{3}$is

$\mathrm{\Delta }{\varphi }^{\prime }=\frac{2\pi }{\lambda }\times \frac{\lambda }{3}=\frac{2\pi }{3}$

Thus, intensity of light corresponding to this phase difference of$\frac{2\pi }{3}$is

$I^{\prime }=4I_0\times {\left(\cos \frac{2\pi }{3}\right)}^2$

$=K\times {\left[-\frac{1}{2}\right]}^2$[from Equation. (i)]

$=\frac{K}{4}$