In YDSE, the source placed symmetrically with | vedclass

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(D) The fringe width $\beta = \frac{\lambda D}{d}$ depends only on the wavelength of light $\lambda,$ the distance between the slits $d,$ and the dist

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

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(D) The fringe width $\beta = \frac{\lambda D}{d}$ depends only on the wavelength of light $\lambda,$ the distance between the slits $d,$ and the distance between the screen and the slits $D.$ Since these parameters remain unchanged,the fringe width 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 $S_2.$ The central bright fringe occurs where the path difference is zero. To compensate for the shorter path to $S_1,$ the point of zero path difference must shift towards the side of the slit that is further from the source,which is the lower slit $S_2.$ Therefore,the entire fringe pattern shifts downwards.

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

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To make the central fringe appear at the centre $O$ of the screen,a mica sheet of refractive index $\mu = 1.5$ is introduced in front of one of the slits. Choose the correct statement.

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