How Does Light's Wave Nature Affect Double Slit Experiment Results?

[Dx]

Hello,

In a double slit experiment the slit separation is 2.0mm and two wavelengths 750nm and 900nm illuminate the slits. A screen placed 2m from the slits. at what distance from the central maximum on the screen will a bright fringe from one pattern first conincide with a bright fringe from the other?

my formula?
d = sin[the] / (m [lamb]

Is my answer correct 6.0mm

Thanks!
Dx

 

[schwarzchildradius]

at what distance from the central maximum on the screen will a bright fringe from one pattern first conincide with a bright fringe from the other?

d=2E-3meters, λ1=750E-9m, λ2=900E-9m, D=2m (D being adjacent leg of right triangle, y1 and y2 being opposite legs, θ being angle between slits.)
y1 = mλ1D / d = (1)(750E-9m)(2m)/(2E-3)m = 7.5E-4m
y2 = nλ2D / d = (1)(900E-9m)(2m)/(2E-3)m = 9.0E-4m
were n=m=1, WHICH THEY're not.
You can't solve it trigonometrically.
In order to solve it, you need to equate
y1=y2=mλ1D/d = nλ2D/d
Then you just solve for m:
m = (λ2/λ1)n = 1.2n
The first instance of this relation being true is if m=6 and n=5
Now you just solve y1 and y2 above to get...

y1=y2=0.0045m

 

[M98Ranger]

Hello Dx,

Your answer is not quite correct. The formula you provided is the correct one to use, however, the value for "m" should be the same for both wavelengths since we are looking for the first bright fringe that coincides for both patterns. The value for "m" represents the order of the fringe, with m=0 being the central maximum. So, for the first bright fringe to coincide, we need to solve for m=1 for both wavelengths.

Using the given values, we get:

d = sin[the] / (m [lamb])

For 750nm:
d = sin[the] / (1 x 750nm)
d = sin[the] / 750nm

For 900nm:
d = sin[the] / (1 x 900nm)
d = sin[the] / 900nm

Since we are looking for the same distance, we can set these two equations equal to each other and solve for d:

sin[the] / 750nm = sin[the] / 900nm
Cross-multiplying and solving for d, we get:
d = (750nm x 900nm) / 750nm
d = 900nm

Therefore, the first bright fringe from one pattern will coincide with the first bright fringe from the other at a distance of 900nm from the central maximum on the screen.

I hope this helps clarify things. Keep up the good work!