Young's Interference Exp: Min Distance b/w Bright Fringes 540nm & 450nm

  • Thread starter Thread starter DDS
  • Start date Start date
  • Tags Tags
    Young
AI Thread Summary
In a Young's interference experiment with slits separated by 0.132 mm and light wavelengths of 540 nm and 450 nm, the goal is to find the minimum distance from the center of the screen where bright fringes coincide. The correct approach involves setting up equations for the positions of the bright fringes for both wavelengths and equating them to find a common position. The ratio of the fringe orders must be calculated using the wavelengths to determine where the patterns overlap. This requires using the formula for fringe position and solving for the distance. The final solution will provide the minimum distance where the bright lines coincide.
DDS
Messages
171
Reaction score
0
In a Young's interference experiment, the two slits are separated by 0.132 mm, and the incident light includes light of wavelengths λ1 = 540 nm and λ2 = 450 nm. The overlapping interference patterns are formed on a screen 1.47 m from the slits. Calculate the minimum distance from the center of the screen to the point where a bright line of the λ1 light coincides with a bright line of the λ2 light.

I know the formula for the location of a bright fringe is:

Yb=m*lamda*L/d

so do i just calculate the distance of a bright fringe at the 540 wavelenght and then the 450 wavelength and then just the distance b/w them is the answer to the question?
 
Physics news on Phys.org
DDS said:
In a Young's interference experiment, the two slits are separated by 0.132 mm, and the incident light includes light of wavelengths λ1 = 540 nm and λ2 = 450 nm. The overlapping interference patterns are formed on a screen 1.47 m from the slits. Calculate the minimum distance from the center of the screen to the point where a bright line of the λ1 light coincides with a bright line of the λ2 light.

I know the formula for the location of a bright fringe is:

Yb=m*lamda*L/d

so do i just calculate the distance of a bright fringe at the 540 wavelenght and then the 450 wavelength and then just the distance b/w them is the answer to the question?

No. The question asks you for the location where both patterns have bright lines that coincide, i.e., happen in the same place. You have two equations for the two different wavelengths that must yield the same value for Y. How is that possible?
 
So do i set up each equation with there corresponding wavelenght , equate them and find y?

What process do i need to undertake to get my asnwer
 
DDS said:
So do i set up each equation with there corresponding wavelenght , equate them and find y?

What process do i need to undertake to get my asnwer

The process you need to undertake is for you to try to solve the problem the way you think it needs to be solved. You obviously have an idea how to do it. Follow your idea and see where it leads.
 
i followed it the way i think and it lead to the wrong asnwer that's why i am asking for help
 
DDS said:
i followed it the way i think and it lead to the wrong asnwer that's why i am asking for help

Show us what you did.
 
DDS said:
i followed it the way i think and it lead to the wrong asnwer that's why i am asking for help

Ok, what must be true to have both diffraction gratings to coincide with each other? Don't mind the distances yet...
 
its a ration between

m1/m2=lada2/lamda1

find the ratio plug it back into eitehr distance formula and bingo answer time
 

Similar threads

Replies
3
Views
1K
Replies
3
Views
780
Replies
6
Views
4K
Replies
8
Views
2K
Replies
1
Views
2K
Replies
4
Views
2K
Replies
4
Views
5K
Back
Top