In Young's double-slit experiment using the monochromatic

Solution:

Let $I_{1}$ and $I_{2}$ be the intensity of the two light waves. Their resultant intensities can be obtained as:

$I^{\prime}=I_{1}+I_{2}+2 \sqrt{I_{1} I_{2}} \cos \phi$

Where,

$\phi=$ Phase difference between the two waves

For monochromatic light waves:

$I_{1}=I_{2}$

Therefore $I^{\prime}=I_{1}+I_{2}+2 \sqrt{I_{1} I_{2}} \cos \phi$

$=2 I_{1}+2 I_{1} \cos \phi$

Phase difference $=\frac{2 \pi}{\lambda} \times$ Path difference

Since path difference $=\lambda$, Phase difference, $\phi=2 \pi$ and $\mathrm{I}^{\prime}=\mathrm{K}$ [Given]

Therefore $I_{1}=\frac{K}{4} \cdots \cdots \cdots \cdots$ (i)

When path difference $=\frac{\lambda}{3}$

Phase difference, $\phi=\frac{2 \pi}{3}$

Hence, resultant intensity:

$I_{g}^{\prime}=I_{1}+I_{1}+2 \sqrt{I_{1} I_{1}} \cos \frac{2 \pi}{3}$

$=2 I_{1}+2 I_{1}\left(-\frac{1}{2}\right)$

Using equation (i), we can write:

$I_{g}=I_{1}=\frac{K}{4}$

Hence, the intensity of light at a point where the path difference is $\frac{\lambda}{3}$ is $\frac{K}{4}$ units.