The wavelength of the light used in Young's d | vedclass

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(D) In Young's double slit experiment,the phase difference $\phi$ is related to the path difference $\Delta x$ by the formula: $\phi = \frac{2\pi}{\la

Vedclass · Accepted Answer

$0.75$

Vedclass · Answer

(D) In Young's double slit experiment,the phase difference $\phi$ is related to the path difference $\Delta x$ by the formula: $\phi = \frac{2\pi}{\lambda} \Delta x$.Given the path difference $\Delta x = \lambda/6$,the phase difference is $\phi = \frac{2\pi}{\lambda} 	imes \frac{\lambda}{6} = \frac{\pi}{3} = 60^{\circ}$.The intensity at any point is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$. Assuming $I_1 = I_2 = I'$,the maximum intensity is $I_0 = I' + I' + 2\sqrt{I' I'} \cos(0^{\circ}) = 4I'$.Substituting the values for the given point: $I = I' + I' + 2\sqrt{I' I'} \cos(60^{\circ}) = 2I' + 2I'(0.5) = 2I' + I' = 3I'$.Therefore,the ratio is $\frac{I}{I_0} = \frac{3I'}{4I'} = 0.75$.

The wavelength of the light used in Young's double slit experiment is $\lambda$. The intensity at a point on the screen is $I$,where the path difference is $\lambda/6$. If $I_0$ denotes the maximum intensity,then the ratio of $I$ and $I_0$ is

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