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(A) The path difference is given by $\Delta x = \frac{7 \lambda}{4}$.The phase difference $\phi$ is related to the path difference by the formula $\ph

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(A) The path difference is given by $\Delta x = \frac{7 \lambda}{4}$.The phase difference $\phi$ is related to the path difference by the formula $\phi = \frac{2 \pi}{\lambda} \Delta x$.Substituting the value of $\Delta x$: $\phi = \frac{2 \pi}{\lambda} 	imes \frac{7 \lambda}{4} = \frac{7 \pi}{2}$.The intensity $I$ at any point is given by $I = I_{\max} \cos^2\left(\frac{\phi}{2}ight)$.Therefore,the ratio $\frac{I}{I_{\max}} = \cos^2\left(\frac{7 \pi}{2 	imes 2}ight) = \cos^2\left(\frac{7 \pi}{4}ight)$.Using the property $\cos(2 \pi - 	heta) = \cos(	heta)$,we get $\cos^2\left(2 \pi - \frac{\pi}{4}ight) = \cos^2\left(\frac{\pi}{4}ight)$.Since $\cos\left(\frac{\pi}{4}ight) = \frac{1}{\sqrt{2}}$,we have $\left(\frac{1}{\sqrt{2}}ight)^2 = \frac{1}{2}$.

Similar Questions

In a Young's double slit experiment,the ratio of the amplitude of light coming from the slits is $2:1$. The ratio of the maximum to minimum intensity in the interference pattern is:

The distance between two slits in a double slit experiment is $1 \ mm$. The distance between the slits and the screen is $1 \ m$. If the distance of the $10^{th}$ fringe from the central fringe is $5 \ mm$, then the wavelength of light is $... \mathring{A}$.

White light is used to illuminate two slits in a $YDSE$. The separation between the slits is $d$ and the screen is at a distance $D (D >> d)$ from the slits. At a point on the screen directly in front of one of the slits,which of the following wavelengths are missing?

Assertion : No interference pattern is detected when two coherent sources are infinitely close to each other.
Reason : The fringe width is inversely proportional to the distance between the two slits.

Young's experiment demonstrates that . . . . . .

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