From Handout:
1.
a) 2.5 mm
b) wavelength = 385 nm, frequency = 6 x 10¹⁴ Hz
c) Since the wavelength decreases, and the spacing between maxima is proportional to wavelength, the space between the maxima will decrease.

1991B6.
a) 3.9 x 10^-5 m
b) 9.6 x 10^-3 m

Problems:
7. 8.5 x 10^-5m
10. The important thing here is to notice that the ratio d/L is the same for both. You should obtain the equation that 2*lambda_1/y = m*lambda_2/2*y, with y the same for both since the question asks for the wavelength at which a minimum appears at the same location. Solving for a wavelength lambda_2 in the visible spectrum, lambda_2 = 613 nm
20. Solve for theta using sin(theta)= m*lambda/W, and then use the fact that L*tan(theta) = height above the center. The distance is 1.591 meters.
21. Notice that this says the distance between maxima, not minima. The path difference must be half a wavelength more than that required for the minimum. The total angle is 19.5 degrees across the central maximum, so the angle between the central maximum and the first minimum is 9.75 degrees. sin(9.75) = 1.5*lambda/W, so W = 5606 nm.