# In Young’s double-slit experiment using monochromatic light of wavelengthλ,

Question:

In Young’s double-slit experiment using monochromatic light of wavelengthλ, the intensity of light at a point on the screen where path difference is λ, is units. What is the intensity of light at a point where path difference is λ /3?

Solution:

Let I1 and I2 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,

$I_{1}=I_{2}$

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

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

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

Since path difference = λ,

Phase difference, $\phi=2 \pi$

$\therefore I^{\prime}=2 I_{1}+2 I_{1}=4 I_{1}$

Given,

I’ = K

$\therefore I_{1}=\frac{K}{4}$    ...(1)

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

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

Hence, resultant intensity, $I_{R}^{\prime}=I_{1}+I_{1}+2 \sqrt{I_{1} I_{1}} \cos \frac{2 \pi}{3}$

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

Using equation (1), we can write:

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

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