11%) Problem 6: Light of wavelength = 490 nm and intensity I0 = 170 W/m2 is inci

Question

11%) Problem 6: Light of wavelength = 490 nm and intensity I0 = 170 W/m2 is incident upon two narrow slits that are separated by a distance d = 48 m. 25% Part (a) Write an equation for the cosine of half of the phase difference between the waves passing through the two slits in terms of the initial intensity I0 and the resulting intensity I. 25% Part (b) What is the smallest phase angle, in radians, for which the resulting intensity of the light will be I = 65 W/m2? 25% Part (c) Choose the correct formula for the diffraction angle, , relative to the normal in terms of , d, I0, and I. 25% Part (d) What is the smallest angle from the normal , in radians, for which the resulting intensity of the light will be I = 65 W/m2?

Explanation / Answer

lamdba = 490*10^-9 m

Io = 170 W/m^2

d = 48 micro m

from fraunhoffer intensity formula

a. I(theta) = Io*lambda^2*cos^2(pi*d*sin(theta)/lambda)sin^2(pi*b*sin(theta)/lambda)/(pi*b*sin(theta))^2)

b. I = 65 W/m^2

for b = 0

65/170 = cos^2(pi*d*sin(theta)/lambda)

theta = 0.0029379 rad = 0.168 deg

c. formula is

I(theta) = Io*cos^2(pi*d*sin(theta)/lambda)

d. 0.0029379 rad

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