Wavelengths: Length between 2nd-order fringes

In summary, a light of wavelengths 4.80x10^2 nm and 632nm passes through two slits 0.52 mm apart, and we are asked to determine the distance between the second-order fringes on a screen 1.6m away. Using the formulas for dark and bright fringes, we can calculate the values of x1 and x2 for each wavelength. Then, by finding the difference between x1 and x2 for the second-order fringes, we can determine the distance between them. However, it is important to note that the choice of using dark or bright fringes depends on the specific wavelengths being used.
  • #1
okandrea
5
0

Homework Statement



Light of wavelenghs 4.80x10^2 nm and 632nm passes through two slits 0.52 mm apart. How far apart are the second-order fringes on a screen 1.6m away?

λ₁ = 4.80x10^2 nm = 4.80x10^-7m
λ₂ = 6.32x10^-7m
d = 0.52mm = 5.2x10^-4m
n = 2
L = 1.6

Homework Equations



(Maxima/Bright)
x/L = nλ/d
(Minima/Dark)
x/L = (n - 1/2)λ/d

*subscript of X would be n in both cases

△x = | x₁ - x₂ |

The Attempt at a Solution


I wasn't so sure which of the two formulas I would be using because there doesn't seem to be a clear indication as to whether or not it's bright/dark (this was what I mainly struggled with).

I tried using both but I don't understand if either of them are correct. I rearranged for x in both equations (moving the L variable to the right) and repeated it for each wavelength:

(A) Using dark:
x₁ = ((2 - 1/2)(4.80x10^-7)(1.6))/5.2x10^-4
x₁ = 2.2x10^-3m

x₂ = (2 - 1/2)(6.32x10^-7)(1.6))/5.2x10^-4
x₂ = 2.9x10^-3 m

△x = | 2.2x10^-3 - 2.9x10^-3 |
△x = 7.0x10^-4 m

(B) Using bright:
x₁ = ((2)(4.80x10^-7)(1.6))/5.2x10^-4
x₁ = 3.0x10^-3 m

x₂ = ((2)(1.6)(6.32x10^-7))/5.4x10^4
x₂ = 3.9x10^-3 m

△x = | 3.0x10^-3 - 3.9x10^-3 |
△x = 9.0x10^-4 m

They aren't too far off. I don't quite trust the textbook solutions since plenty of wavelength-related solutions were wrong. They did, however, use the formula for wavelengths with dark fringes...
 
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  • #2
If neither "bright" nor "dark" is explicitly mentioned, I would assume "bright", but that's my personal interpretation. Also, for better accuracy, I would first find an algebraic formula for Δx using just symbols and then put in the numbers.
 
Last edited:
  • #3
kuruman said:
I would first find an algebraic formula for Δx using just symbols and then put in the numbers.
Would that mean something like this?
Δx/L = λ/d
Δx = Lλ/d
 
  • #4
okandrea said:
Would that mean something like this?
Δx/L = λ/d
Δx = Lλ/d
Not really, what happened to the ##n## in the expression? You need two expressions, one for each wavelength. It helps being organized.
1. For wavelength 1 you have ##x_1=nL\lambda_1/d##.
2. Write a similar expression for ##x_2##.
3. Find an algebraic expression for the difference ##x_2-x_1## for the second order fringes.
4. Put in the numbers.

On edit: My earlier statement that it doesn't matter if you use dark or bright fringes is incorrect. It does make a difference if the wavelengths are different. I edited that statement.
 

1. What is a wavelength?

A wavelength is a measure of the distance between two consecutive points on a wave that are in phase with each other. It is usually measured in meters or nanometers.

2. How is wavelength related to the 2nd-order fringes?

The 2nd-order fringes refer to the bright or dark bands formed when light waves interfere with each other. The distance between these fringes is directly related to the wavelength of the light used.

3. How is the wavelength of light measured using 2nd-order fringes?

The wavelength of light can be measured by using a diffraction grating or interferometer which produces 2nd-order fringes. By measuring the distance between these fringes and knowing the distance between the light source and the grating, the wavelength can be calculated using the formula λ = d*sinθ, where d is the distance between the grating lines and θ is the angle between the incident light and the fringes.

4. Can the wavelength of light change?

The wavelength of light depends on the medium it is traveling through. In a vacuum, the wavelength of light is constant. However, when light travels through a different medium, such as air or water, its wavelength can change due to the change in the speed of light.

5. What is the significance of measuring wavelengths?

Measuring wavelengths is important in various fields such as optics, astronomy, and telecommunications. It allows us to understand the properties of light and how it behaves in different mediums. It also helps in the development of technologies that use light, such as lasers and fiber optics.

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