An interference pattern is produced by light with a wavelength 600 nm from a dis

Question

An interference pattern is produced by light with a wavelength 600 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.460 mm .

1.If the slits are very narrow, what would be the angular position of the first-order, two-slit, interference maxima?

2.

What would be the angular position of the second-order, two-slit, interference maxima in this case?

2 = radians

C

Let the slits have a width 0.320 mm . In terms of the intensity I0 at the center of the central maximum, what is the intensity at the angular position of 1?

  I0

What is the intensity at the angular position of 2?

Explanation / Answer

in the two slit interference maxima are given by

m*lambda = d*sin A

1.

for m = 1

sin A = m*lambda/d

sin A = 1*600*10^-9/(0.460*10^-3)

A = arcsin (1*600*10^-9/(0.460*10^-3)) = 0.0013043 rad

2.

A = arcsin (2*600*10^-9/(0.460*10^-3)) = 0.0026086 rad.

3.

Intensity is given by

I = Io*cos ^2(phi/2)*[(sin (B/2))/(B/2)]

phi = (2*pi*d*sin A)/lambda

phi = (2*pi*d/lambda)*(m*lambda/d) = 2*pi*m

cos ^2(phi/2) = cos ^2 (m*pi) = 1

for any value of m

B = (2*pi*a*sin A)/lambda

B = (2*pi*a/lambda)*(m*lambda/d) = 2*pi*m*(a/d)

B = 2*pi*m*(0.320/0.460) = m*4.3686

Now for m = 1

I1 = Io*1*[(sin (4.3686/2))/(4.3686/2)]^2

I1 = 0.1401*Io

Now for m = 2

I2 = Io*1*[(sin (4.3686))/(4.3686)]^2

I2 = 0.0464*Io

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