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(B) The fringe width $\beta$ in a Young's double-slit experiment is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the de Broglie wavelengt

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the accelerating voltage is increased and the distance of the screen from the slits is decreased.

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(B) The fringe width $\beta$ in a Young's double-slit experiment is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the de Broglie wavelength,$D$ is the distance of the screen from the slits,and $d$ is the distance between the slits.For electrons accelerated through a potential $V$,the de Broglie wavelength is $\lambda = \frac{h}{\sqrt{2meV}}$.Substituting this into the fringe width formula,we get $\beta = \frac{hD}{d\sqrt{2meV}}$.To decrease $\beta$,we need to increase the denominator ($d$ or $V$) or decrease the numerator $(D)$.Option $B$ states that the accelerating voltage $V$ is increased and the distance $D$ is decreased. Both actions contribute to a decrease in $\beta$.

Imagine a Young's double-slit interference experiment performed with waves associated with fast-moving electrons produced from an electron gun. The distance between successive maxima (fringe width) will decrease if:

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In a double-slit experiment,the distance between the slits is $d$. The screen is at a distance $D$ from the slits. If a bright fringe is formed opposite to one of the slits,its order $n$ is:

In a Young's double slit experiment,the source illuminating the slits is changed from blue to violet. The width of the fringes

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