The intensity at the maximum in a Young's dou | vedclass

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(C) Given: $d = 5\lambda$,$D = 10d$,and the position in front of one slit is $y = \frac{d}{2}$.The path difference $\Delta x$ at position $y$ is given

Vedclass · Accepted Answer

$\frac{I_0}{2}$

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(C) Given: $d = 5\lambda$,$D = 10d$,and the position in front of one slit is $y = \frac{d}{2}$.The path difference $\Delta x$ at position $y$ is given by $\Delta x = d \sin 	heta \approx d 	an 	heta = d \left( \frac{y}{D} ight)$.Substituting the values: $\Delta x = d \left( \frac{d/2}{10d} ight) = \frac{d}{20} = \frac{5\lambda}{20} = \frac{\lambda}{4}$.The phase difference $\phi$ is $\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \left( \frac{\lambda}{4} ight) = \frac{\pi}{2}$.In a Young's double slit experiment,the resultant intensity $I_y$ is given by $I_y = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.Since $I_1 = I_2 = I$,we have $I_{max} = I + I + 2\sqrt{I^2} = 4I = I_0$,so $I = \frac{I_0}{4}$.Substituting $\phi = \frac{\pi}{2}$ into the intensity formula:$I_y = I + I + 2I \cos(\frac{\pi}{2}) = 2I + 0 = 2I$.Since $I = \frac{I_0}{4}$,the intensity $I_y = 2 \left( \frac{I_0}{4} ight) = \frac{I_0}{2}$.

The intensity at the maximum in a Young's double slit experiment is $I_0$. The distance between two slits is $d = 5\lambda$,where $\lambda$ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance $D = 10d$?

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