This document discusses the math behind the double slit interference simulation on the PhET website. It begins by explaining how the simulation allows users to manipulate different factors and visualize their effects. However, it notes that a deeper understanding requires analyzing the equations. It then presents three example questions working through the calculations for double slit interference with different wavelengths of light and slit separations. The calculations show that increasing the wavelength or decreasing the slit separation decreases the number of observable fringes by increasing the angular spacing between them. Working through the math provides a more thorough understanding of how to apply double slit concepts.

1. Double slit interference
Pre-reading in depth
Delving into the math behind what
happened in the phet simulation

2. 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.

3. 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

4. • The values that are given are:
– D=2590nm = 2.59x10-6m
– Λ= 430nm for blue and 670nm for orange
– Slit separation d=1749nm =1.749x10-6m
First we find sinθ for blue
• 𝑑 =
𝑚𝜆
𝑠𝑖𝑛𝜃
• 1.749𝑥10−6
𝑚 =
1𝑥4.3𝑥10−7 𝑚
𝑠𝑖𝑛𝜃
• 𝑠𝑖𝑛𝜃 =
4.3𝑥10−7 𝑚
1.749𝑥10−6 𝑚
• 𝑠𝑖𝑛𝜃 = 0.246
• sin−1
0.246 = 14.23˚

5. • Now we need to find θ for orange light:
• 𝑑 =
𝑚𝜆
𝑠𝑖𝑛𝜃
• 1.749𝑥10−6
𝑚 =
1𝑥6.7𝑥10−7 𝑚
𝑠𝑖𝑛𝜃
• 𝑠𝑖𝑛𝜃 =
6.7𝑥10−7 𝑚
1.749𝑥10−6 𝑚
• 𝑠𝑖𝑛𝜃 = 0.383
• sin−1
0.383 = 22.52˚
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

6. • Using the values we can find the distance between fringe 0 and 1
for blue and orange light:
• Blue light:
• 𝑡𝑎𝑛𝜃 =
𝑦
𝐷
• tan 14.23 =
𝑦
2.59𝑥10−6 𝑚
• tan 14.23 𝑥 2.59𝑥10−6
𝑚 = 𝑦 = 6.56𝑥10−7
𝑚
• Orange light:
• 𝑡𝑎𝑛𝜃 =
𝑦
𝐷
• tan 22.62 =
𝑦
2.59𝑥10−6 𝑚
• tan 22.52 𝑥 2.59𝑥10−6 𝑚 = 𝑦 = 1.073𝑥10−6 𝑚
• 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

7. • Next we use what was found to find the
maximum m value, using 90 degrees for sin:
• 𝑚 =
𝑑𝑠𝑖𝑛𝜃
𝜆
=
1.749𝑥10−6 𝑚 𝑥 𝑠𝑖𝑛 90
4.3𝑥10−7 𝑚
= 4.06
• Truncating to lowest integere means there are
four fringes.
• Using the wavelength of orange:
• 𝑚 =
𝑑𝑠𝑖𝑛𝜃
𝜆
=
1.749𝑥10−6 𝑚 𝑥 𝑠𝑖𝑛 90
6.7𝑥10−7 𝑚
= 2.6
• This means that only two fringes will appear,
thus the amount of fringes decreases

8. 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 4.0𝑥10−7 𝑚
• 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)

9. • With a d value of 1.0𝑥10−6 𝑚 we get
• 𝑑 =
𝑚𝜆
𝑠𝑖𝑛𝜃
• 1.0𝑥10−6 𝑚 =
1𝑥4.0𝑥10−7 𝑚
𝑠𝑖𝑛𝜃
• 𝑠𝑖𝑛𝜃 =
4.0𝑥10−7 𝑚
1.0𝑥10−6 𝑚
• 𝑠𝑖𝑛𝜃 = 0.4
• sin−1
0.4 = 23.57˚
When the value is doubled to 2.0𝑥10−6 𝑚 we get:
• 𝑑 =
𝑚𝜆
𝑠𝑖𝑛𝜃
• 2.0𝑥10−6 𝑚 =
1𝑥4.0𝑥10−7 𝑚
𝑠𝑖𝑛𝜃
• 𝑠𝑖𝑛𝜃 =
4.0𝑥10−7 𝑚
2.0𝑥10−6 𝑚
• 𝑠𝑖𝑛𝜃 = 0.2
• sin−1 0.2 = 11.54˚

10. • Using the θ values to calculate the distance between fringes yields:
First value of θ for d= 1.0𝑥10−6 𝑚
• 𝑡𝑎𝑛𝜃 =
𝑦
𝐷
• tan 23.57 =
𝑦
2.59𝑥10−6 𝑚
• tan 23.57 𝑥 2.59𝑥10−6
𝑚 = 𝑦 = 1.13𝑥10−6
𝑚
Second value of θ for d= 2.0𝑥10−6
𝑚
• 𝑡𝑎𝑛𝜃 =
𝑦
𝐷
• tan 11.54 =
𝑦
2.59𝑥10−6 𝑚
• tan 11.54 𝑥 2.59𝑥10−6 𝑚 = 𝑦 = 5.28𝑥10−7 𝑚
• As can be seen, the distance between fringes is inversely
proportional to the distance between slits.

11. • Using the new d values into the fringe amount
equation we get:
• 𝑚 =
𝑑𝑠𝑖𝑛𝜃
𝜆
=
1.0𝑥10−6 𝑚 𝑥 𝑠𝑖𝑛 90
4.0𝑥10−7 𝑚
= 2.5
• 𝑚 =
𝑑𝑠𝑖𝑛𝜃
𝜆
=
2.0𝑥10−6 𝑚 𝑥 𝑠𝑖𝑛 90
4.0𝑥10−7 𝑚
= 5
• Thus it can be observed that the amount of
fringes increases proportionally to the
distance between slits

12. 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:
– 𝑡𝑎𝑛𝜃 =
𝑦
𝐷
– D x 𝑡𝑎𝑛𝜃=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

13. Works cited
• http://phet.colorado.edu/en/simulation/wave
-interference
• http://www.webassign.net/web/Student/Assi
gnment-Responses/last?dep=11176487 for
pre-reading questions