White light is used to illuminate two slits i | vedclass

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(D) In a $YDSE$ setup,the path difference $\Delta x$ at a point on the screen at a distance $y$ from the central axis is given by $\Delta x = \frac{yd

Vedclass · Accepted Answer

Both $(A)$ and $(C)$

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(D) In a $YDSE$ setup,the path difference $\Delta x$ at a point on the screen at a distance $y$ from the central axis is given by $\Delta x = \frac{yd}{D}$.For a point directly in front of one of the slits,the distance from the central axis is $y = \frac{d}{2}$.Substituting this into the path difference formula: $\Delta x = \frac{(d/2)d}{D} = \frac{d^2}{2D}$.For destructive interference (missing wavelengths),the path difference must be an odd multiple of half the wavelength: $\Delta x = (2m + 1) \frac{\lambda}{2}$,where $m = 0, 1, 2, ...$.Equating the two expressions: $\frac{d^2}{2D} = (2m + 1) \frac{\lambda}{2}$.Solving for $\lambda$: $\lambda = \frac{d^2}{(2m + 1)D}$.For $m = 0$,$\lambda = \frac{d^2}{D}$.For $m = 1$,$\lambda = \frac{d^2}{3D}$.Thus,both wavelengths $\frac{d^2}{D}$ and $\frac{d^2}{3D}$ are missing.

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?

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