A double-slit apparatus is held 1.2m from a screen. When red light(600nm) is sent through the double slit, the interference pattern shows a?
a distance of 12.5cm between the 1st and 10th dark fringes. What is the separation of the slits?
What will be the difference in path length for the waves traveling from each slit to the tenth nodal line?
- nyphdinmdLv 75 months agoFavorite Answer
y = m*wl*d/a where m = 1, 2, 3,..., wl = wavelength, d = distance from slits to screen, a = distance between slits
delta_y = y(m) - y(n) = wl*d/a*(m - n)
So for a given everything else ---> a = wl*d*(m-n)/delta_y = 600x10^-9(m)*1.2(m)*(10 - 1)/(0.125 (m)) = 5.184x10^-5 m
- ∅Lv 75 months ago
you should ask your teacher. unless you want to pay us what you pay him...