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

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In summary, you would need to have the distances between the two gratings to be the same, and then you would need to solve for the ratio between the two distances.
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DDS
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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?
 
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  • #2
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?
 
  • #3
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
 
  • #4
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.
 
  • #5
i followed it the way i think and it lead to the wrong asnwer that's why i am asking for help
 
  • #6
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.
 
  • #7
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...
 
  • #8
its a ration between

m1/m2=lada2/lamda1

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

1. What is Young's interference experiment?

Young's interference experiment is a classic experiment in optics that demonstrates the wave nature of light. It involves passing a beam of light through a narrow slit and then through two closely spaced slits, creating an interference pattern of bright and dark fringes on a screen.

2. How is the minimum distance between bright fringes determined in Young's interference experiment?

The minimum distance between bright fringes, also known as the fringe spacing, is determined by the wavelength of the light used and the distance between the two slits. It can be calculated using the equation d*sin(θ) = mλ, where d is the slit spacing, θ is the angle between the bright fringes, m is the order of the fringe, and λ is the wavelength of the light.

3. What is the significance of using light of two different wavelengths in Young's interference experiment?

Using light of two different wavelengths in Young's interference experiment allows us to observe the interference pattern and measure the fringe spacing for each wavelength. This helps us to better understand the wave nature of light and the principles of interference.

4. Why is it important to use monochromatic light in Young's interference experiment?

Monochromatic light, which consists of a single wavelength, is used in Young's interference experiment because it produces a more distinct and easily measurable interference pattern. Using light with multiple wavelengths can result in overlapping and distorted fringes, making it difficult to accurately measure the fringe spacing.

5. How can the results of Young's interference experiment be applied in real-world situations?

The principles of Young's interference experiment have numerous applications in various fields, such as optics, microscopy, and interferometry. It is used to measure the wavelength of light, study the properties of materials, and even in the development of technologies like holography and fiber optics.

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