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(D) In Young's Double Slit Experiment,the fringe width $\beta$ is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the

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All of the above.

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(D) In Young's Double Slit Experiment,the fringe width $\beta$ is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance between the slits and the screen,and $d$ is the slit separation.$1$. Since $\beta \propto D$,the fringe width changes in direct proportion to $D$. Thus,option $(B)$ is correct.$2$. The angular fringe width $	heta$ is given by $	heta = \frac{\beta}{D} = \frac{\lambda}{d}$. Since $	heta$ depends only on $\lambda$ and $d$,it does not change when $D$ is varied. Thus,option $(A)$ is correct.$3$. The central maximum occurs at the point where the path difference is zero. This point is equidistant from both slits,which is the point $O$ on the screen. Changing $D$ does not shift this point. Thus,option $(C)$ is correct.Since all statements $(A)$,$(B)$,and $(C)$ are correct,the correct option is $(D)$.

The figure shows a schematic diagram of Young's Double Slit Experiment. If the distance $D$ between the slits and the screen is varied,choose the correct statement$(s)$.

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