The single slit

When light passes through a single slit whose width w is on the order of the wavelength of the light, then we can observe a single slit diffraction pattern on a screen that is a distance L >> w away from the slit.  The intensity is a function of angle.  Huygens' principle tells us that each part of the slit can be thought of as an emitter of waves.  All these waves interfere to produce the diffraction pattern.  Where crest meets crest we have constructive interference and where crest meets trough we have destructive interference.

Very far from a point source the wave fronts are essentially plane waves.  This is called the Fraunhofer regime, and the diffraction pattern is called Fraunhofer diffraction.  The positions of all maxima (constructive interference) and minima (destructive interference) in the Fraunhofer diffraction pattern can be calculated fairly easily.
The positions of the first minimum in the Fraunhofer diffraction pattern from a single slit can be found from the following simple arguments.

• Consider a slit of width w, as shown in the diagram on the right.  A plane wave is incident from the bottom and all points oscillate in phase inside the slit.
• For light leaving the slit in a particular direction defined by the angle θ, we may have destructive interference between the ray at the right edge (ray 1) and the middle ray (ray 7).  To arrive at a distant screen perpendicular to the direction of propagation of the rays, the rays coming from different points inside the slit have to travel different distances.  They have a different optical path length.  If ray 7 has to travel an extra distance of one-half wavelength (λ/2) compared to ray 1, then ray 1 and ray 7 destructively interfere.  Crest meets trough.
• The optical path length (OPL) of a light ray traveling from point A to point B is defined as c times the time it takes the ray to travel from A to B.  In free space the optical path length is just the distance d between the points.  In a transparent medium with index of refraction n, it equals n times this distance, OPL = nd,  because the light moves with speed c/n.

If the optical path length of two rays differs by λ/2, the two rays interfere destructively.  For ray 1 and ray 7 to be half a wavelength out of phase we need

(w/2)sinθ = λ/2 or w sinθ = λ.

But from geometry, if these two rays interfere destructively, so do rays 2 and 8, 3 and 8, and 6 and 10, 5 and 11, and 6 and 12.
In effect, light from one half of the opening interferes destructively and cancels out light from the other half.

Destructive interference produces the dark fringes.  Dark fringes in the diffraction pattern of a single slit are found at angles θ for which

w sinθ = mλ,

where m is an integer, m = 1, 2, 3, ... .   For the first dark fringe we have w sinθ = λ.

When w is smaller than λ , the equation w sinθ = λ has no solution and no dark fringes are produced.

Problem:

When a monochromatic light source shines through a 0.2 mm wide slit onto a screen 3.5 m away, the first dark band in the pattern appears 9.1 mm from the center of the bright band.  What is the wavelength of the light?

Solution:

• Reasoning:
  Dark fringes in the diffraction pattern of a single slit are found at angles θ for which w sinθ = mλ, where m is an integer, m = 1, 2, 3, ... .   For the first dark fringe we have w sinθ = λ.  Here we are asked to solve this equation for λ.
• Details of the calculation:
  z = 9.1 mm = 9.1*10^-3 m.
  L = 3.5 m.
  w = 0.2 mm = 2*10^-4 m.
  L >> z, therefore sinθ ~ z/L and λ = zw/(mL).
  λ = (9.1*10^-3 m)(2*10^-4 m)/(3.5 m).
  λ = 5.2*10^-7 m = 520 nm.

Problem:

Consider a single slit diffraction pattern for a slit width w.  It is observed that for light of wavelength 400 nm the angle between the first minimum and the central maximum is 4*10^-3 radians.  What is the value of w?

Solution:

• Reasoning:
  Dark fringes in the diffraction pattern of a single slit are found at angles θ for which w sinθ = mλ, where m is an integer, m = 1, 2, 3, ... .   For the first dark fringe we have w sinθ = λ.  Here we are asked to solve this equation for w.
• Details of the calculation:
  First minimum:  w sinθ = λ, w = (400 nm)/sin(4*10^-3 radians) = 1*10^-4 m.