In YDSE,the source S placed symmetrically wit | vedclass

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(D) The fringe width $\beta$ is given by the formula $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance between the s

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the fringe width will remain same but the fringe pattern will shift down.

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(D) The fringe width $\beta$ is given by the formula $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance between the slits and the screen,and $d$ is the distance between the two slits. Since $\lambda$,$D$,and $d$ remain unchanged,the fringe width $\beta$ remains constant.When the source $S$ is moved closer to the upper slit $S_1$,the path length from $S$ to $S_1$ becomes shorter than the path length from $S$ to $S_2$. To maintain a zero path difference at the screen,the point on the screen must be closer to $S_2$ than to $S_1$. This means the central bright fringe (where path difference is zero) shifts downwards. Consequently,the entire fringe pattern shifts downwards.

In $YDSE$,the source $S$ placed symmetrically with respect to the slits $S_1$ and $S_2$ is now moved parallel to the plane of the slits so that it is closer to the upper slit $S_1$,as shown. Then,

Similar Questions

$A$ 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:

The minimum thickness of a mica sheet having refractive index $\mu = 3/2$,which should be placed in front of one of the slits in $YDSE$ to reduce the intensity at the center of the screen to half of the maximum intensity,is:

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