Intensity in Young's double slit experiment problem.

IrAlien
Messages
19
Reaction score
0
Hello all, I have a problem that requires some knowledge about Young's double slit experiment.

The question is as follows:
Imagine performing Young's experiment with one slit wider than the other so that the intensity passing the wider slit is four times as great as that passing the narrower slit. Assume both slits are still 'thin' and act as line sources and show that I = Inaught.(5+4cos(phi))

I do not know where to start... can anyone give me some tips?

Do I use I=4Inaught.[cos(phi/2)^2] ??

Thanks,
Levi
 
Physics news on Phys.org
You have two sources of electromagnetic waves on a plane. The sources are separated by a given distance. They travel through a given distance through space and illuminate second plane.

What you need to do is find the total amplitude of the two electric fields on the illuminated plane, and then find the irradiance.
 
Ok thank you for the help. I _think_ I've got it.

Levi
 
ummm..cant you use Cosine rule to solve this??
 
I'm not sure.

I probably wouldn't use it, but it might be useful.
 
Last edited:
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...

Similar threads

Back
Top