Young's double slit experiment fringes

AI Thread Summary
Young's double slit experiment raises questions about the number of illuminated fringes formed, particularly why they aren't infinite despite the absence of screen limitations. The key factor is the angle of the observed maxima, which restricts the number of fringes based on the equation dsin(α)=nλ, where α must be less than 90 degrees. Additionally, the angle at which the incident beam strikes the slits does affect the interference pattern, as it alters the path length differences between light waves. Understanding these principles clarifies the behavior of light as secondary sources in the experiment. Overall, the geometry of the setup and the properties of light dictate the observable outcomes.
aim1732
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I have some questions regarding Young's double slit experiment.

1.The first concerns the no. of possible fringes that will be formed.
Sample question is as:
Slit Separation = 1mm and wavelength of light is 5250 A.
No. of illuminated fringes=?
The basic problem is that I don't understand why the no. of fringes is not infinite as there is no limitation on the extent of the screen and light can interfere anywhere.

2.Then there is this thing about the incident beam being at an angle to the normal of the plane of the slits. Does it really matter if the light makes an angle? Anyhow the points illuminated are going to behave as secondary sources. Although I think it will but i can not put my thoughts together.
Any help would be appreciated.
 
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aim1732 said:
I have some questions regarding Young's double slit experiment.

1.The first concerns the no. of possible fringes that will be formed.
Sample question is as:
Slit Separation = 1mm and wavelength of light is 5250 A.
No. of illuminated fringes=?
The basic problem is that I don't understand why the no. of fringes is not infinite as there is no limitation on the extent of the screen and light can interfere anywhere.

The size of the screen is not the limiting factor, the angle of observed maxima is. Solve n in dsin(α)=nλ for α<90 degrees (you have all the other variables).
 
Thanks. I should have seen that before.
Part 2?
 
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