Blue light of wavelength 400 nm is incident on a four-slit barrier where the sli
ID: 3279269 • Letter: B
Question
Blue light of wavelength 400 nm is incident on a four-slit barrier where the slits are evenly separated by distances of 1.40 mu m each. The slits are equally illuminated so that each produces the same amplitude. A screen is placed 80.0 cm from the slits. (a) Calculate the maximum value of m for this setup. (b) Find the locations on the screen for the m = 1 and m = 2 peaks. (c) At what locations on the screen between these peaks will there be zeros in the intensity? (d) For each of the zeros that you have found draw a phasor diagram. (e) For the location of x = 40.0 cm on the screen calculate the relative intensity of the light.Explanation / Answer
3. (a) lambda = 400 x 10^-9 m
a = 1.40 x 10^-6 m
D = 0.80 m
sin(theta_m) = m *lambda / a
for maximum m, theta_m = 90 deg
sin90 = m (400 x 10^-9) / (1.40 x 10^-6)
m = 3.5
so maximum value = 3
(B) y = m lambda D / a
y1 = 400x 10^-9 x 0.80 / (1.40 x 10^-6)
y1 = 0.23 m ........Ans
y2 = (2 x 400 x 10^-9 x 0.80) / (1.40 x 10^-6)
y2 =0.46 m ..........Ans
(C) for m = 1.5
y = (1.5)(400 x 10^-9) (0.80) / (1.40 x 10^-6)
y = 0.34 m
(e) at y = 40 cm = 0.40 m
phi = (2 pi / lambda) a sin(theta)
y = m * lambda * D / a
y = sin(theta) D
sin(theta) = 0.40 / 0.80 = 0.5
phi = 3.5 pi
relative intensity = (cos(phi/2))^2 = 0.032