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# As the footnote on above page indicates, the separation, $\Delta x$, between two adjacent bright areas or fringes in the double slit interference pattern is determined by the distance, $a$, between the slits, the distance, $S$, between the plane of the slits and the screen on which the pattern is projected, and the wavelength, $\lambda$, of light used in the experiment. Specifically, the relationship among these variables is given by:$\Delta x=\left(\frac{S}{a}\right) \lambda .$Based on this equation, for a given slit spacing and screen distance, for which color of light, red or blue, will the fringe spacing be greater? For a given slit spacing and wavelength, if the screen distance, $S$, is increased, what happens to the fringe spacing? If the separation between the slits is decreased, for a given value of $\lambda$ and $S$, how will $\Delta x$ change? Suppose that a beam of red light from a He-Ne laser $(\lambda=633 \mathrm{~nm})$ strikes a screen containing two narrow slits separated by $0.200 \mathrm{~mm}$. The fringe pattern is projected on a white screen located $1.00 \mathrm{~m}$ away. Find the distance in millimeters between adjacent bright areas near the center of the interference pattern that develops. As mentioned in the text, the relationship given above could be used to calculate the wavelength of light in an experiment in which the other three quantities are measured.

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In order to solve this problem, we will answer the questions one by one. Firstly, since the double-slit experiment is described by the equation

$\Delta x=\frac{S}{a}\cdot\lambda$

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