A monochromatic beam of light falls on a YDSE | vedclass

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(D) In a $YDSE$ setup,when light is incident at an angle $	heta$,the initial path difference at point $O$ is $d \sin 	heta$ (where $d$ is the slit s

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Anywhere depending on the angle $	heta$,thickness of plate $t$,and refractive index of glass $\mu$

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(D) In a $YDSE$ setup,when light is incident at an angle $	heta$,the initial path difference at point $O$ is $d \sin 	heta$ (where $d$ is the slit separation). When a glass plate of thickness $t$ and refractive index $\mu$ is placed in front of slit $S_2$,it introduces an additional path difference of $(\mu - 1)t$. The net path difference at point $O$ becomes $\Delta x = d \sin 	heta - (\mu - 1)t$. The central bright fringe is formed where the net path difference is zero. $1$. If $d \sin 	heta = (\mu - 1)t$,the path difference is zero at $O$,so the central fringe is at $O$. $2$. If $d \sin 	heta > (\mu - 1)t$,the path difference is positive,meaning the central fringe shifts above $O$. $3$. If $d \sin 	heta Thus,the position depends on the values of $	heta$,$t$,and $\mu$.

$A$ monochromatic beam of light falls on a $YDSE$ apparatus at an angle $\theta$ as shown in the figure. $A$ thin sheet of glass of thickness $t$ and refractive index $\mu$ is inserted in front of the lower slit $S_2$. The central bright fringe (path difference $= 0$) will be obtained:

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$A$ thin transparent plate of thickness $t$ and refractive index $\mu$ is introduced in the path of one of the two interfering waves. The path difference between these waves is changed by half the wavelength of light $\lambda$. The thickness of the plate is

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