Discussion Forum

In young's double-slit experiment, the intensity of tight at a point on the screen where the path difference is λ, is I, λ being the wavelength of light used. The intensity at a point where the path difference is λ/4 will be

Answer:

Explanation:

For path difference λ, phase difference = $2\pi$  ($Q = \left(\frac{2\pi}{\lambda}x=\frac{2\pi}{\lambda}\lambda=2\pi\right)$)

$I = I_{0}+I_{0}+2I_{0}\cos2\pi$

$I = 4I_{0} (\therefore\cos 2\pi = 1)$

For $x= \frac{\lambda}{4}$, Phase difference = $\frac{\pi}{2}$

$I' = I_{1}+I_{2}+2\sqrt{I_{1}}\sqrt{I_{2}}\cos\pi/2.$

$If = I_{1} = I_{2} = I_{0}$ then I' = $2I_{0}$ = 2.I/4 = I/2