Young's experiment is performed with light of wavelength 502 nm from excited helium atoms. Fringes are measured carefully on a screen 1.00 m away from the double slit, and the center of the 20th fringe (not counting the central bright fringe) is found to be 11.4 mm from the center of the central bright fringe.
a. What is the separation of the two slits?

0.88 mm

Explanation:

Given

Wavelength of the atoms, λ = 502 nm = 502*10^-9 m

Radius of the screen away from the double slit, r = 1 m

We know that Y(20) = 11.4*10^-3 m

d = (20 * R * λ) / Y(20)

d = (20 * 1 * 502*10^-9)/11.4*10^-3

d = 1*10^-5 / 11.4*10^-3

d = 0.88 mm

Therefore, the distance of separation between the two slits is 0.88 mm

i think the answer is a, hope this is right, that would be my best guess.

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