Solving C-28-3 w/o PhET!

*Mathematical Approach*

By the unfortunate student who needs to make up their LO badly

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

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 thatmλ = distance that a waves travels Where m is a strictly positive integer in 2D interference

How do we know if constructive or destructiveInterference occur at a particular point?

Skip the next few slide if you completely understand 2D interference!

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 Now we use this equation to get the following:

• = C

If C = integer + ½ , then destructive interference occursIf C = integer , then constructive interference occursIf you don’t understand why, revisit these concepts!

Note that the two waves of interest have the same wavelength and frequency

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

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 screen2) Increase/Reduce wavelength of light3) Increase/Decrease spacing of slits

Moving the slits toward/away from Screen• What this means: we are increasing or decreasing D • What we are not changing: d, θ,

→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.

Increasing or decreasing Wavelength of Light• What this means: we are increasing or decreasing the wavelength ) • 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 . 𝜆If this is NOT clear, please refer to the first few slides aboutDouble-Slit interference and 2D interference.

The next slide will only make sense only if you understand the basics of double-slit and 2D interference.

Increasing or decreasing (Cont’d)𝜆• 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 = m z𝜆1= dsin(θ)

Increasing or decreasing (Cont’d)𝜆• 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

Increasing or decreasing spacing of slits• What this means: we are increasing or decreasing d• What we are not changing: D, , m (assuming m=1)𝜆𝜆m = 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.

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

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