Problem 4: In a multiple-slit interference experiment, 15 narrow slits are equally spaced 0.25 mm apart and illuminated with light of wavelength 570 nm. The interference pattern is projected on a screen 2.4 m away. 25% Part (a) What is the angular position, in degrees, of the third principal maximum? A=0.391 ✓ Correct! 25% Part (b) What is the angular position, in degrees, of the fourth principal maximum? 0 = 0.522 ✓ Correct! 25% Part (c) If y is the position along the screen measured from the center of the pattern, then what is the distance, Ay between the third and fourth principal maxima on the screen? Recall that the screen is located 2.4 m from the barrier with the slits. Ay=0.00548 ✓ Correct! * 25% Part (d) In an N-slit interference pattern, there are N - 2 secondary maxima and N - 1 dark minima between each pair of sequential principal maxima. The dark minima are equally spaced between the centers of the pair of principal maxima. Calculate the width, 8y, of the third principal maximum. dy= 0.000000712 Deductions 1% Potential Grade Summary mi 99%

c) If y is the position along the screen measured from the center of the pattern, then the distance, Ay between the third and fourth principal maxima on the screen = 0.00548m

d) In an N-slit interference pattern, there are N - 2 secondary maxima and N - 1 dark minima between each pair of sequential principal maxima. The dark minima are equally spaced between the centers of the pair of principal maxima. The width is = dy = 0.000000712 m

The vector total of the amplitudes of the individual waves is equal to the consequent amplitude at a location where two or more waves of the same sort are incident, according to the principle of superposition of waves. When the crests of two waves of the same frequency collide at the same spot, constructive interference happens.

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