Question

# The intensity at the central maxima in Young's double slit experiment set-up is $$I_0$$. Show that the intensity at a point where the path difference is $$\lambda/3$$ is $$I_0/4$$.

Solution

## As the resultant intensity at a point,$$I=I_1+I_2+2\sqrt{I_1I_2}\cos\phi$$When the path difference=$$\lambda$$phase difference=$$0°$$$$\therefore I_R=I+I+2\sqrt {I\times I}\cos 0°=2I+2\sqrt{I^2}\times1=2I+2I=4I ........(i)$$When the path difference =$$\dfrac {\lambda} 3$$, phase difference, $$\phi=\dfrac {2\pi}3$$$$\therefore I_R'=I+I+2\sqrt{I\times I}\cos(120°)=2I+2\sqrt{I^2}\times\left(-\dfrac 12\right)=2I-\dfrac {2I}2=I$$$$\therefore I'=\dfrac {I_0}4$$Physics

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