yeah

1. Solving C-28-3 w/o PhET!
*Mathematical Approach*
By the unfortunate student who needs to make up their LO badly

2. Investigating Double-Slit Systems
• Figure 1 shows all the information you need to fully
understand the Double-Slit system
• Waves that enter through S1 travel less distance to
point P than the waves that enter through S2
*look at the line segment joining S1 and P and S2 and P (figure 1) *
Using trigonometry, you can figure out that dsin θ is
equal to the extra distance that waves ( that go
through S2 and not S1) need to travel to get to point P
Figure 1
-Slit 1 = S1
-Slit 2 = S2
-The two slits are separated by
the distance d
-θ = angle between d and line
connecting S1 and point Q

3. Refresher: Interference in 2D
Forget about double slit systems for a second, and refer back to 2D
interference of waves with the same wavelength and same frequency
From 2D interference, we know that
mλ = distance that a waves travels
Where m is a strictly positive integer in 2D interference
How do we know if constructive or destructive
Interference occur at a particular point?
Skip the next few slide if you completely understand 2D interference!

4. Refresher: Interference in 2D Cont’d
• Distance1 = distance between point A and wave source 1
• Distance2 = distance between point A and wave source 2
• Distance 2 > Distance 1
• We rearrange this equation mλ = distance to get
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝜆
= 𝑚
Now we use this equation to get the following:
•
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 2
𝜆
−
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 1
𝜆
= C
If C = integer + ½ , then destructive interference occurs
If C = integer , then constructive interference occurs
If you don’t understand why, revisit these concepts!
Note that the two waves of interest have
the same wavelength and frequency

5. Checkpoint C-28-3
• What does this actually mean?
Basically, the fringes are too
close together and you want
to increase the distance
between the shaded areas,
which is called “y” in the
figure below
View of the screen
This is a bird’s eye view of
the experimental set up

6. Checkpoint C-28-3 Cont’d
• What we want to do is basically increase the
y distance between point P and point O
Note that these waves are in phase and of same
wavelength regardless of which slit it comes out from as it
is from the SAME SOURCE.
How can we do that?
Options given:
1) Move the slits toward/away from the screen
2) Increase/Reduce wavelength of light
3) Increase/Decrease spacing of slits

7. Moving the slits toward/away from Screen
• What this means: we are increasing or decreasing D
• What we are not changing: d, θ, 𝜆
• 𝐻𝑜𝑤 𝑑𝑜 𝑤𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑑𝑜𝑒𝑠𝑛′
𝑡 𝑐ℎ𝑎𝑛𝑔𝑒?
→We look at the key equation (28-5); Does the angle depend on D?
• By having the same angle as before, by increasing or
decreasing D, we create a similar triangle.
Increasing D would give us a larger similar triangle,
meaning larger y.

8. Increasing or decreasing Wavelength of Light
• What this means: we are increasing or decreasing the wavelength (𝜆)
• What we are not changing: D, d, m (assuming m=1)
(d = distance between the slits, which is a constant that can only be changed by making new slits that are farther
apart from each other; even though it is included in the equation, it does not change.
We assume m=1 because it makes it easier to understand….I’m going to exploit the fact that m has to be a strictly
positive integer!!)
There has to be constructive interference at point P because there exists a
fringe. For a point to have constructive interference, wave from S2 has to
travel a certain distance which is a multiple of its 𝜆.
If this is NOT clear, please refer to the first few slides about
Double-Slit interference and 2D interference.
The next slide will only make sense only if you understand the basics of
double-slit and 2D interference.

9. Increasing or decreasing 𝜆 (Cont’d)
• Let’s make a hypothetical situation!
• We are using light with wavelength 𝜆1 for a Double-Slit experiment and
using 𝜆1 creates a fringe at point P
• For 𝜆1, extra distance = dsin(θ) = 𝜆1 ; meaning m=1
• Now let’s use light with wavelength 𝜆2 for the exact same experimental set
up.
The following = true:
• 𝜆2> 𝜆1 and 𝜆2=z𝜆1 , where z = positive constant bigger than 1
Therefore! Extra distance for 𝜆2 = mz𝜆1= dsin(θ)

10. Increasing or decreasing 𝜆 (Cont’d)
• mz𝜆1= dsin(θ)
→ if 𝜆1 was increased by a factor of z, then the original m should decrease by z as
well in order to keep the equality of both sides of the equation …
but we know m = strictly positive integer and 1 is the lowest strictly positive integer.
• Now we have to increase the other side by z to
make both side of the equation equal!
• As mentioned d is CONSTANT,
• so the only other variable that we can change is the angle
• With a bigger angle (between 0 degrees and 90 degrees)…
…..take a look at the diagram in the bottom right corner
• Pink line = constructive interference with 𝜆2
• Black line = constructive interference with 𝜆1
Longer wavelength = Bigger angle = longer distance
= higher y value

11. Increasing or decreasing spacing of slits
• What this means: we are increasing or decreasing d
• What we are not changing: D, 𝜆, m (assuming m=1)
𝜆m = CONSTANT = dsin(θ)
→ If we increase d, then sin(θ) has to decrease.
→ If we decrease d, then sin(θ) has to increase.
From our investigation of wavelength, we know BIGGER angles lead to higher
y value.
Decreasing d = increase in the angle = increased y value.

12. Answer
• Move the slits farther away from the screen
• Increase the wavelength
• Decrease the d, meaning decrease the distance between the two slits
• The answer is D

13. Thanks for reading!
• I hope my LO helped you out 
• Leave a thumb up if it helped you.