In a YDSE apparatus,if we use white light,the | vedclass

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(D) In $YDSE$ with white light,the path difference for the central fringe is $0$ for all wavelengths. Since all wavelengths overlap at the center,the

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All of the above.

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(D) In $YDSE$ with white light,the path difference for the central fringe is $0$ for all wavelengths. Since all wavelengths overlap at the center,the central fringe appears white.As we move away from the center,the path difference $\Delta x = d \sin 	heta$ increases. For a given point,different wavelengths satisfy the condition for constructive interference at slightly different positions. Since violet light has the shortest wavelength,the first order maximum for violet appears closest to the central fringe.Because different wavelengths have different positions for their maxima and minima,the fringes overlap significantly,preventing the formation of a completely dark fringe (where intensity would be zero for all wavelengths simultaneously).Therefore,all the given statements are correct.

In a $YDSE$ apparatus,if we use white light,then:

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