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Understanding Fraunhofer Diffraction
Learning Goal: To understand the derivations of, and be able to use, the equations for Fraunhofer diffraction.

Diffraction is a general term for interference effects related to edges or apertures. Diffraction is more familiar in waves with longer wavlengths than those of light. For example, diffraction is what causes sound to bend around corners or spread as it passes through a doorway. Water waves spread as they pass between rocks near a rugged coast because of diffraction. Two different regimes for diffraction are usually identified: Fresnel and Fraunhofer.

Fresnel diffraction is the regime in which the diffracted waves are observed close (as compared to the size of the object causing the diffraction) to the place where they are diffracted. Fresnel diffraction is usually very complicated to work with. The other regime, Fraunhofer diffraction, is much easier to deal with. Fraunhofer diffraction applies to situations in which the diffracted waves are observed far from the point of diffraction. This allows a number of simplifying approximations to be used, reducing diffraction to a very manageable problem.

An important case of Fraunhofer diffraction is the pattern formed by light shining through a thin slit onto a distant screen (see the figure).

Notice that if the light from the top of the slit and the light from the bottom of the slit arrive at a point P on the distant screen with a phase difference of 2\pi, then the electric field vectors E_vec of the light from each part of the slit will cancel completely, resulting in a dark fringe. To understand this phenomenon, picture a phasor diagram for this scenerio (as show in the figure).

A phasor diagram consists of vectors (phasors) with magnitude proportional to the magnitude of the electric field of light from a certain point in the slit. The angle of each vector is equal to the phase of the light from that point. These vectors are added together, and the resultant vector gives the net electric field due to light from all points in the slit. In the situation described above, since the magnitude of the electric field vectors is the same for light from any part of the slit and the angle of the phasors changes continuously from 0 to 2\pi, the phasors will make a complete circle, starting and ending at the origin. The distance from the origin to the endpoint of the phasor path (also the origin) is zero, and so the magnitude of the electric field at point P is zero.

Part A
One reason that Fraunhofer diffraction is relatively easy to deal with is that the large distance from the slit to the screen means that the light paths will be essentially parallel.

Therefore, the distance d marked in the figure is the entire path-length difference between light from the top of the slit and light from the bottom of the slit. What is the value of d?

Hint A.1  

Hint not displayed

Express your answer in terms of the slit width a and the angle theta shown in the figure.
ANSWER:
  d =  a{\cdot}{\sin}\left(theta\right) 
Part B
As described in the problem introduction, a criterion for a dark band to appear at point P is that the phase difference between light arriving at point P from the top of the slit and light arriving at point P from the bottom of the slit equal 2\pi. What length of path difference d will give a phase difference of 2\pi?
Express your answer in terms of the wavelength lambda.
ANSWER:
  d =  {\lambda} 
Combining your answers from Parts A and B gives the criterion for a dark band in the diffraction pattern as a\sin(\theta)=\lambda.
Part C

Consider the phasor diagram from the introduction. The magnitude of the electric field at a point will equal zero as long as the endpoint for the phasor diagram is the origin. Thus, a point with a phasor diagram that goes around a circle twice, for example, ending at the origin, will be another location for a dark band. This idea can be used to modify the equation for the location of a dark band by introducing a variable m:

a\sin(\theta)=m\lambda.

What is the complete set of values of m for which this equation gives criteria for dark bands?

ANSWER:
1, 2, 3,\dots
 0, 1, 2, 3,\dots
\pm1, \pm2, \pm3, \dots
 0, \pm1, \pm2, \pm3,\dots
any rational number
The value m=0 corresponds to \theta=0, which is the center of the diffraction pattern. The center of the diffraction pattern is a bright band. To see why, notice that if the phase difference from top to bottom is zero, then the phasor diagram will just be a straight line segment pointing away from the origin. This gives the maximum possible intensity in the diffraction pattern.
Part D
What are the angles for the two dark bands closest to the central maximum.
Express your answers in terms of a and lambda. Separate the two angles with a comma.
ANSWER:
 arc{\sin}\left(\frac{{\lambda}}{a}\right),-arc{\sin}\left(\frac{{\lambda}}{a}\right) 
Part E

The equation for the angles to dark bands is valid for any angle from -90^\circ to 90^\circ. In practice, the bright bands at large angles are usually so dim that the diffraction pattern appearing on a screen is invisible for such angles. For small angles, it is easy to find the distance y_m from the center of the diffraction pattern to the dark band on the screen corresponding to a particular value of m.

For small angles, \sin(\theta) \approx \tan(\theta). Since \tan(\theta)=y/x, the small-angle approximation yields \sin(\theta)=y/x. By solving the dark-band criterion, you obtain \sin(\theta)=m\lambda/a. Setting the two expressions for \sin(\theta) equal gives the formula for the position (i.e., distance from the center of the diffraction pattern) of dark bands:

\frac{y}{x}=\frac{m\lambda}{a},

or equivalently,

y=\frac{mx\lambda}{a}.

Assuming that the angle between them is small, what is the distance Deltay between the two dark bands closest to the center of the diffraction pattern?

Express your answer in terms of x, a, and lambda.
ANSWER:
  Deltay =  \frac{2{\cdot}x{\cdot}{\lambda}}{a} 
Part F
Suppose that light from a laser with wavelength 633 \rm nm is incident on a thin slit of width 0.500 \rm mm. If the diffracted light projects onto a screen at distance 1.50 \rm m, what is the distance y_2 from the center of the diffraction pattern to the dark band with m=2?
Express your answer in millimeters to two significant figures.
ANSWER:
  y_2 =  3.80   \rm mm
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