In Young’s double slit experiment, the intensity on the screen at a point where path difference $\lambda$ is $K$. What will be the intensity at the point where path difference is $\lambda/4$

Question

Dr. Abubacker · Accepted Answer

(b) By using phase difference $\phi = \frac{2\pi}{\lambda}(\Delta)$
For path difference $\lambda$, phase difference $\phi_{1} = 2\pi$ and for path difference $\lambda/4$, phase difference $\phi_{2} = \pi/2$
Also by using $I = 4I_{0}\cos^{2}{\frac{\phi}{2} \Rightarrow \frac{I_{1}}{I_{2}} = \frac{\cos^{2}{(\phi_{1}/2)}}{\cos^{2}{(\phi_{2}/2)}}}$
$\Rightarrow \frac{K}{I_{2}} = \frac{\cos^{2}{(2\pi/2)}}{\cos^{2}\left( \frac{\pi/2}{2} \right)} = \frac{1}{1/2} \Rightarrow I_{2} = \frac{K}{2}$

Q. In Young’s double slit experiment, the intensity on the screen at a point where path difference $\lambda$ is $K$. What will be the intensity at the point where path difference is $\lambda/4$

Explanation

Reviewed by Dr. Abubacker

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