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

Question

Dr. Abubacker · Accepted Answer

(b) When path difference is $\lambda,\ I_{\max} = 4I = K$
When path difference is $\frac{\lambda}{4}$, phase difference, $\phi = \pi/2$
$	herefore\ \ I_{R} = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}}\cos\phi$
$= I + I = 2I = \frac{K}{2}$

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

Explanation

Reviewed by Dr. Abubacker

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