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(B) In a double-slit diffraction experiment,the interference pattern is modulated by the diffraction envelope. The number of interference fringes $n$

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Ratio of slit separation to slit width

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(B) In a double-slit diffraction experiment,the interference pattern is modulated by the diffraction envelope. The number of interference fringes $n$ within the central diffraction maximum is given by the ratio of the width of the central diffraction maximum to the fringe width of the interference pattern.The angular width of the central diffraction maximum is $\Delta 	heta_{diff} = \frac{2\lambda}{a}$,where $a$ is the slit width.The angular fringe width of the interference pattern is $\beta_{	heta} = \frac{\lambda}{d}$,where $d$ is the distance between the slits.The number of interference fringes $n$ is the ratio of these two widths:$n = \frac{2\lambda / a}{\lambda / d} = \frac{2d}{a}$.Thus,the number of interference fringes depends on the ratio of the slit separation $d$ to the slit width $a$.

On which factor does the number of interference fringes in a given width of a diffraction peak depend?

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