Double slit interferencePre-reading in depth

Delving into the math behind what happened in the phet simulation

In the phet simulation at http://phet.colorado.edu/en/simulation/wave-interference. On the light tab of the simulation, several factors can be manipulated to observe their effect on double-slit interference of light of different wavelengths. The simulation gives a marvelous visual representation and allows adept-level understanding of the concept via demonstration. However, simply understanding the concept may not be sufficient, and understanding the math and minutiae behind what was observed in the simulation can give a more in depth understanding of how to apply these concepts to double-slit interference questions.

First concept question• Choose the light tab, select one light and two slits. Set slit

width to on tick above 0nm, maximum light source amplitude, barrier location to 2590( no units stated, will use nm), slit separation to one tick below 1750(will use nm)

• What happens in the simulation if you move from blue light to orange light?– The distance between fringes increases– The amount of fringes decreases– The distance from the central maximum to the first fringe increases

• The values that are given are:– D=2590nm = 2.59x10-6m– Λ= 430nm for blue and 670nm for orange– Slit separation d=1749nm =1.749x10-6mFirst we find sinθ for blue

• Now we need to find θ for orange light:

This may not seem like much but such an infinitesimal difference can mean a great deal when dealing with phenomena on such a minute and precise scale

• Using the values we can find the distance between fringe 0 and 1 for blue and orange light:

• Blue light:

• Orange light:

• Notice that the distance between fringes increases as you move from blue to orange light, naturally this would mean that there are fewer fringes since the screen is the same height, but let’s prove it mathematically

• Next we use what was found to find the maximum m value, using 90 degrees for sin:

• Truncating to lowest integere means there are four fringes.

• Using the wavelength of orange:

• This means that only two fringes will appear, thus the amount of fringes decreases

Second question

• All previous variables are the same but violet light is used and the distance between slits is increased.

• The wavelength of violet light is • What happens if you increase the distance between two

slits.– The distance between fringes decreases– The distance from the central maximum to the first fringe

decreases– More fringes appear on the screen

• First we can calculate the respective θ for increasing slit separation values(d)

• With a d value of we get

When the value is doubled to we get:

• Using the θ values to calculate the distance between fringes yields:

First value of θ for d=

Second value of θ for d=

• As can be seen, the distance between fringes is inversely proportional to the distance between slits.

• Using the new d values into the fringe amount equation we get:

• Thus it can be observed that the amount of fringes increases proportionally to the distance between slits

Third question• Same as #2 but this time the D value (distance between slits and

screen) is changed• What happens if the barrier gets closer to the screen(D value

decreases)?– More fringes appear– Distance between fringes decreases– Distance from central maximum descreases

• Looking at the relationships:

– D x =y– The distance y between each fringe increases with an increasing D value,

thus there are fewer fringes if the D value is higher (if the barrier is farther from the screen) and the opposite is true

Works cited

• http://phet.colorado.edu/en/simulation/wave-interference

• http://www.webassign.net/web/Student/Assignment-Responses/last?dep=11176487 for pre-reading questions