# Young's Double Slit with 2 Wavelengths variant

• JohnGaltis
In summary, the question asks for the angles, represented by theta, where bright fringes are formed on a screen for both the wavelengths of 500nm and 600nm in a Double Slit experiment using a point light source. The separation of the two slits is 1mm and the equation used to find the angles is d sin θ= ΔL = +- mλ (bright fringes). There are an infinite amount of fringes, but it ends at 90 degrees. A possible answer for the angles is given by the formula θ = arcsin (mλ/d) with m = 0, ±1, ±2, ... and for m << (d/λ) the values are θ =

## Homework Statement

A point light source is used in a Double Slit experiment. The light source contains two wavelengths(500nm and 600nm).

Separation of the two slits d=1mm. Two sets of interference fringes are formed on a screen. Find the angles θ where bright fringes are formed for both the wavelengths on the screen.

## Homework Equations

d sin θ= ΔL = +- mλ (bright fringes)

## The Attempt at a Solution

The question asked for angles. From the fringes created from the 500nm and 600nm waves. How many though, I don't know. So the first thing I tried to do was find out how many fringes there will be since it would give me the values of "m" and hence, the angles involved in each value of "m".

And here, I have no idea how.

Assume m = 0, 1, 2, 3, 4, 5, ... etcetera !
Ignore single slit difffraction pattern (no info given -- assume narrow enought).

BvU said:
Assume m = 0, 1, 2, 3, 4, 5, ... etcetera !
Ignore single slit difffraction pattern (no info given -- assume narrow enought).

Hello, BvU, thanks for replying. But I wouldn't know "m" goes how high?

If I use m=0, m=1, m=2, I will get 6 angles, 3 from each wavelength. If I use m=3 too, I get more angles as the answer. So how many values of m should I use? The question didn't specify.

I mean, technically there are an infinite amount of fringes right? And hence, angles.

JohnGaltis said:
technically there are an infinite amount of fringes right
Nope. It stops by the time ##\theta = {\pi\over 2}## .

But I think an answer in the form of ##\theta = \arcsin \left (m\lambda\over d\right )## with ##m = 0, \pm 1, \pm 2, ... ## should be acceptable. If you then complete with: for ## m << {d\over \lambda} ## the values are ##\theta = 0, ... ## and fill in##m = 0, \pm 1, \pm 2, ... ## for both ##\lambda## you are covered on all sides ( unless this is a computerized exercise ?).

BvU said:
Nope. It stops by the time ##\theta = {\pi\over 2}## .

But I think an answer in the form of ##\theta = \arcsin \left (m\lambda\over d\right )## with ##m = 0, \pm 1, \pm 2, ... ## should be acceptable. If you then complete with: for ## m << {d\over \lambda} ## the values are ##\theta = 0, ... ## and fill in##m = 0, \pm 1, \pm 2, ... ## for both ##\lambda## you are covered on all sides ( unless this is a computerized exercise ?).

Oh yeah sheesh, it ends at 90 degrees. And yes, that would work. Thanks for the clarification, BvU.