Yonug double-slit-experiment homework

[burgerkin]

Homework Statement

In Yonug double-slit-experiment, the net I (intensity) is 1/4 of the intensity from each source, at an angle of 29° off the center-axis. There is no minima between this point and the central--maximum. The source is apart for 420 nm. What is the wavelength of the (monochromatic) light?

Homework Equations

\lambda= yD/mLI am not sure which equation to use. I am very confused whenever intensity is involved in a problem.

The Attempt at a Solution

 

[ehild]

See this place.

http://www.uAlberta.ca/~pogosyan/teaching/PHYS_130/FALL_2010/lectures/lect36/lecture36.html

If you scroll down you will find a formula for two slits.

ehild

 

[burgerkin]

To get wavelength we normally use \lambda=\frac{}{}dsin(\theta)/n

so if there is no minima in between which means that n=1, first order.

Can i just use the above formula to calculate the wavelength?

I am just confused about the intensity part. What does it mean by saying net intensity is 1/4 of individual intensity? Is net intensity different from maximum intensity?

\lambda=420nmxsin(29)/1=204nm

it is a wrong answer,tho. so Why can't I calculate like that? what went wrong?
is it because the angle is not a small angle?

but then we can get wavelength by using the intensity formula, but i do not know the relation between I and I max...
I am very confused

 

[ehild]

The resultant intensity of the interfering waves is neither maximum nor minimum at the given angle, at 29°. You have to find the phase difference between the waves which produces 1/4 of the intensity of a single wave. That phase difference is equal to 2pi*Dsin(theta)/lambda where D is the distance of the sources -those are the two slits.

ehild

 

[burgerkin]

Thanks a lot!

I see that if I find the phase difference then I can find wavelength of the light.

So the \varphi that makes the net intensity 1/4 of the single intensity is at \pi/8?? Is that correct?I am not sure if i know how to find the phase difference. when I add 2 sine waves, at 29 degree, I need to get an resultant that gives me a phase difference? I am sorry if I sound confusing...

oh no, I don't think it is at 29 degree, I am messing it up.

but 29 degree is related to path difference, so it is related to phase difference, but how do I use the intensity to find the phase difference?

 

[ehild]

The phase difference δ of the diffracted waves is related to the path difference Dsinθ as
δ=2π/λ*(Dsinθ).
Two waves of the same frequency and polarization and traveling in the same direction is equivalent with a single wave: that is the essence of interference.
These two waves come from the two slits. As they originate by diffraction from the same incident light beam, the amplitudes are equal. The sum of the two diffracted waves is equal to a single wave with amplitude E₁ and phase constant φ :

E₀sin(ωt-2π/λ*x)+E₀sin(ωt-2π/λ*x+δ)=E₁sin(ωt-2π/λ*x+φ )

You can use the sum rule sin(α+β)=sin(α)cos(β)+cos(α)sin(β) to expand the sine of sum of angles, and collect the terms with sin(ωt-2π/λ*x) and cos(ωt-2π/λ*x):

sin(ωt-2π/λ*x)(E₀(1+cosδ)-E₁cosφ)+cos(ωt-2π/λ*x)(E₀sinδ-E₁sinφ)=0

The equation must hold for every t and x and this can happen only when both
E₀(1+cosδ)-E₁cosφ=0
and
E₀sinδ-E₁sinφ=0

that is,

E₀(1+cosδ)=E₁cosφ

and

E₀sinδ=E₁sinφ

Take the square of both equations, add them and you get the square of the amplitude of the new wave:

2E₀²(1+cosδ)=E₁²

The intensity of the wave is proportional to the square of the amplitude so you can rewrite the equation for the intensities in terms of the phase difference between the waves:

I₁=2I₀(1+cosδ)

ehild

 

[burgerkin]

Thanks so very much! It was very helpful. I was confused with I max and I net. So I kept on using wrong equations. Thanks!