In a double-slit interference experiment, the slit separation is 2.28 microm, th

Question

In a double-slit interference experiment, the slit separation is 2.28 microm, the light wavelength is 532 nm, and the separation between the slits and the screen is 4.48 m. (a) What is the angle between the center and the third side bright fringe? If we decrease the light frequency to 92.5% of its initial value, (b) does the third side bright fringe move along the screen toward or away from the pattern's center and(c) how far does it move?

Got a which is 44.427 degrees but I need (c) and it is not .254

Explanation / Answer

Note that the fringe spacing is directly proportional to the wavelength.

Thus, if we decrease the wavelength, fringe spacing will also decrease.

Hence, the third bright fringe MOVES TOWARD the pattern's center. [PART B]

*****************

The fringe spacing is

fringe spacing = wavelength * distance / slit separation

Thus, as the new old wavelength is 532 nm,

fringe spacing = 1.0453 m

Thus, the third bright fringe is thrice as far from the center,

distance of third bright fringe = 3.136 m

Thus, the new distance is just 92.5% of this,

new distance of third bright fringe = 2.9008 m

Thus, the distance it moved is

= 3.136 - 2.9008 m

= 0.2352 m   [ANSWER]

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