Problem: Light of wavelength 632.8 nm is incident normally on a system of 4 narr

Question

Problem:

Light of wavelength 632.8 nm is incident normally on a system of 4 narrow slits each slit is 0.04 mm wide, and the center of the each slit is 0.12 mm from the centre of the adjacent slit. An intensity pattern is observed on a distant screen. Three of the slits are blocked, so that light shines through only one slit. Taking the direction from one slit system to the centre of the fringe pattern 0 degree ,what is the angle to the first point of zero intensity? A second slit adjacent to the first, is now uncovered. How many visible interference fringes are there within the central diffraction maximum? Now all four slits arc uncovered. What is the angle to the first point of zero intensity in the resulting pattern? Sketch a graph illustrating the intensity on the screen as a function of angle for the entire region of the central maximum of the diffraction pattern. Your sketch must show the correct number of interference maxima.

Explanation / Answer

a)

sin(theta) = lamda/a = 632.8e-9/0.04e-3 = 0.01582

==> theta = 0.906457 = 0.906 degrees

b)

d sin(theta) = m lamda

a sin(theta) = lamda

therefore:

==> (d sin(theta))/(a sin(theta)) = (m lamda)/(lamda)

==> d/a = m

==> 0.12/0.04 = m

==> m = 3

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