Blue light of wavelength 470 nm is used to illuminate a pair of narrow slits tha
ID: 1450702 • Letter: B
Question
Blue light of wavelength 470 nm is used to illuminate a pair of narrow slits that are 0.020 mm and 1.60 m from a screen. What is the distance on the screen between the central maximum and the second-order minimum (dark spot)? (First find the angle theta of the second-order minimum.) Circle one word from each underlined pair. If you wanted a narrower interference pattern (i.e. you wanted to decrease the angle and distance you found in part (a)), you could use yellow/violet light instead of blue light, or you could increase/decrease the distance between the slits.Explanation / Answer
= 470 nm = 470 x 10^-9 m
D =1.6 m
a = 0.020 mm = 0.020 x 10^-3 m
y = mD/a ---(sin(theta) = y/D and m = 2)
y = (2 x 470 x 10^-9 x 1.6) /(0.020 x 10^-3)
y = 75.2*10^-3
y = 75.2 mm