In a Young's double slit experiment,the path | vedclass

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(B) The path difference is given by $\Delta x = \frac{\lambda}{8}$.The phase difference $\phi$ is related to the path difference by $\phi = \frac{2\pi

Vedclass · Accepted Answer

$0.85$

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(B) The path difference is given by $\Delta x = \frac{\lambda}{8}$.The phase difference $\phi$ is related to the path difference by $\phi = \frac{2\pi}{\lambda} \Delta x$.Substituting the value of $\Delta x$,we get $\phi = \frac{2\pi}{\lambda} 	imes \frac{\lambda}{8} = \frac{\pi}{4}$.The resultant intensity $I$ at any point is given by $I = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos \phi = 2I_0(1 + \cos \phi)$,where $I_0$ is the intensity of each individual wave.At the centre of a bright fringe,the phase difference is $0$,so the maximum intensity $I_{max} = 2I_0(1 + \cos 0) = 4I_0$.At the given point,the intensity $I = 2I_0(1 + \cos(\pi/4)) = 2I_0(1 + \frac{1}{\sqrt{2}})$.The ratio is $\frac{I}{I_{max}} = \frac{2I_0(1 + 1/\sqrt{2})}{4I_0} = \frac{1 + 0.707}{2} = \frac{1.707}{2} \approx 0.85$.

In a Young's double slit experiment,the path difference,at a certain point on the screen,between two interfering waves is $1/8^{th}$ of the wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to:

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