Blue light of wavelength 400 nm is incident on a four-slit barrier where the sli

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

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