Interference+and+Diffraction

= Interference =

Interference happens when waves interfere and add together constructively or destructively. When light waves interfere, they create bright or dark fringes. In order for waves to interfere they must have identical wavelengths and be coherent (the waves emitted must stay at a constant phase). = = =Double Slit Interference=

Thomas Young devised an experiment where he shined a light source through two slits. The waves interfere after passing through the slits, and create bright and dark fringes on the screen, showing constructive and destructive interference of the waves. When the waves interfere constructively, it creates a bright fringe, known as maximum. When they interfere destructively, it creates a dark fringe, known as a minimum. Because the path difference of the rays is //dsinθ//, constructive interference only happens when //** dsinθ=mλ **//. //m// is known as the order number, and is an integer such as 0, 1, 2, ... The central bright fringe is known as the zeroth-order maximum, and occurs when //m//=0. The next maximum above or below the center occurs when //m//=±1, and is known as the first-order maximum. The next maximum after that occurs when //m//=±2.

When //dsinθ// coincides with half of a wavelength, destructive interference happens. Therefore, destructive interference occurs when //dsinθ=(m+1/2)λ//.

This experiment allows the estimation of distance between fringes because //L// is significantly largely than //d.// //θ// is so small, that //sin////θ//≈//tanθ.// Using this, we can determine the distance of bright and dark fringes from the zeroth-order maximum.

// Bright fringe distance: // // Dark fringe distance: //

z=distance of fringe from zeroth-order maximum λ=wavelength of light L=distance of screen from slits d=distance between slits m=order number (0, ±1, ±2, ±3, ...)

=Thin Film Interference=

Thin film interference occurs when light is reflected off of a thin film such as a soap bubble or a thin layer of oil on water. Light waves interfere constructively or destructively when light waves reflecting from the top surface of the thin film interfere with the waves reflecting from the bottom of the film.



Recall that the index of refraction in a surface in a substance is equivalent to λ (wavelength) of light in a vacuum divided by λ of light in the substance. We also know that a wave undergoes a 180° phase change on reflection when n 2 (coating) > n 1 (air), but not when n 2 (coating) > n 1.

Applying these rules leads to the derived formula for constructive and destructive interference in thin film interference.

// Constructive Interference // // Destructive Interference //

n=index of refraction of thin film t=thickness of film m=integer showing a point of interference (0, 1, 2, 3, ...)


 * These equations are true when there is only one phase reversal.** It is only true if the substance above and below the film both have either higher indexes of refraction than the film, or both have lower indexes than the film. If the coating were to have an index higher than both glass and air, the equations would hold true.
 * If the film is placed between a substance with a higher index and a substance with a lower index, the interference equations switch.** So if the coating were to have a lower index than glass, but i higher index than air, it's destructive interference equation would be 2nt=(m+1/2)λ.

= Diffraction =

Diffraction occurs when light passes through a small opening. It creates a pattern with a larger center line of light surrounded by dimmer fringes on either side, alternating with dark fringes.

=Single Slit Diffraction=

The width of a slit actually has significance to the behavior of light waves. Because each portion of light passing through an opening acts as a waves, portions of light interfere with each other. The fringes are shown to occur at certain angles in accordance to the width of the slit and the wavelength of the light.

// Constructive Interference //

// Destructive Interference //

a=width of slit λ=wavelength of light m=order number (0, 1, 2, 3, ...) θ=angle of dark fringe from center =Diffraction Grating=

Diffraction grating occurs when passing light through a series of parallel scratches that act like slits. The light behaves much in the same way as in double slit interference, creating maximums at point along the screen. The condition for a maxima can be calculated from angle and distance between lines.

// Maxima in Diffraction Grating //

d=distance between grating λ=wavelength of light m=order number (0, 1, 2, 3, ...) θ=angle of maxima from center

= Examples =

Example 1
A laser with wavelength of 600nm passes through two slits which are 0.2mm apart. How far away is the first bright fringe from the center on a screen 10 m away from the slits?

Example 2
A light with a wavelength of 550nm is reflected on a glass surface that is covered by a thin coating. The glass's index of refraction is 1.5, and the coating has an index of refraction of 1.3. Find the minimal thickness necessary for //destructive// interference to occur.

Example 3
A light is projected through a slit with width //a.// A dark fringe occurs 7m away from the center light on a screen 15m away. The wavelength of the light is 600nm. Find //a//.

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Solution for Example 1
As this is a double slit experiment, you will be using some variation of //z=λLm/d//.0

First thing to do is figure out if you are looking for a light fringe/maximum/constructive interference, or a dark fringe/minimum/destructive interference. This example is clearly a light fringe, so we will use the equation for bright fringe distance.

Next you want to be sure to get the correct units. 600nm= 6*10^(-7)m. 0.2mm=2*10^(-4)m.

Since //d// is the distance between the two slits, wavelength and //L// are given, and we know //m// is 1 (because we are looking for the first bright fringe), we can plug in numbers to find the answer.

z=(6*10^(-7)m)*(10m)*(1) / (2*10^(-4)m) z=3*10^(-2)m or //3cm//.

**Solution for Example 2**
The first step is determining the number of phase reversal. **Figure out whether or not both mediums surrounding the coating have greater or lesser index's of refraction.** in this case the air on one side has a lower index than the coating, but the glass has a higher index. Thus, there have been 2 phase reversals, and the equations for interference must be flipped.

Determine whether the destructive or constructive equation is called for. In this case, the destructive is need. And since there are 2 phase reversals, the equation for destructive interference is //2nt=(m+1/2)λ.//

The order number //m// can be determined to be 1 because the question asks for the minimal thickness for interference. You could convert units to meters, but because there is only one measurable unit given, the answer can be left with nano-meter units.

2(1.3)t=(1+1/2)(550nm) t=(3/2)(550nm) / (2.6) t=317.3nm

Solution for Example 3
First, determine the equation. When finding the width of a slit in a single slit diffraction problem, use either //sinθ=mλ/a// or //sinθ=(m+1/2)λ/a//. Because you have a destructive interference, you use // sinθ=mλ/a .//

Next, determine what is known. //m// and //λ// are given (//m//=1; //λ//=600nm). //a// and //θ// are still unknown.

//θ// can be determined using the information given. Using geometry, it is know that tan//θ=//opposite/adjacent lines in a triangle. The opposite side from the angle is given (7m), and the adjacent line is given (15m). Therefore, //θ//=arctan(7m/15m)=25°.

The angle can then be plugged into the formula to find //a//.

sin(25 °)=1(600nm)/ // a // // a= // 600nm/(sin25 °) // a // =1419nm