What is the Correct Wavelength of Light in an Interference Experiment?

In summary, the conversation discusses the use of monochromatic light to create an interference pattern on a screen placed parallel to two narrow slits. The distance between the slit openings and the screen is given, as well as the distance between adjacent dark interference fringes. The question asks for the wavelength of the light, with one person getting an answer of 490 nanometers and the textbook giving an answer of 540 nanometers. The conversation then guides the person to use the formula dsinθ = (n+1/2)λ to calculate theta and find the correct answer. The person also questions why they cannot use the formula X/L = lambda/d with three variables given. In summary, the correct answer is not provided but the conversation
  • #1
stphillips
7
0

Homework Statement


Monochromatic light from a point source illuminates two parallel, narrow slits. The centres of the slip openings are 0.8mm apart. An interference pattern forms on a screen placed parallel to the plane of the slits and 49 cm away. The distance between two adjacent dark interference fringes is 0.30 mm.
a) Calculate the wavelength of the light.

Thanks

Homework Equations



I used the formula X/L = lambda/d


The Attempt at a Solution


I got 490 nanometers, the book got 540. Whats the correct answer?
 
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  • #2
If you haven't done paraxial approximations you should be using

dsinθ = (n+1/2)λ where θ=x/l for destructive interference

you know d you know n and x is the distance from the central maximum to every successive light fringe
 
  • #3
I don't know theta in that eq'n though =/
I have 3 variables and i have an eq'n i just don't know why I am getting the wrong answer
 
  • #4
You have enough data to calculate theta
 
  • #5
Okay so what is the right answer then =/ I got an answer it's just different from the textbook and i don't know why
 
  • #6
we cannot tell you how the answer, only to guide you to it. it says that the two dark fringes are .3mm apart. so this would be n=0 since these are the 0 order dark fringes.
What is the distance from one dark fringe to the central maximum?

To find theta, you have the adjacent and you can find the opposite. After that you have everything you need
 
  • #7
Why can't I use the formula X/L = lambda/d
I have 3 of those variables its just when i solve i don't get the right answer :S
I shouldn't have to solve for theta to solve this question

Also, this was homework due like a week ago, now I just want to know the right answer so I know if I'm doing something wrong or if the book is.
 

1. What is the definition of wavelength of light?

The wavelength of light is the distance between two consecutive peaks or troughs in a light wave. It is typically measured in nanometers (nm) or meters (m).

2. How does the wavelength of light affect its color?

The wavelength of light is directly related to its color. Shorter wavelengths appear blue or violet, while longer wavelengths appear red. The visible spectrum ranges from approximately 400nm (violet) to 700nm (red).

3. What is the relationship between wavelength and frequency of light?

The wavelength and frequency of light are inversely proportional. This means that as the wavelength increases, the frequency decreases and vice versa. This relationship is described by the equation: c = λv, where c is the speed of light, λ is the wavelength, and v is the frequency.

4. How is the wavelength of light measured?

The wavelength of light can be measured using various techniques, including diffraction grating, spectrophotometry, and interferometry. These methods involve passing light through a medium or using specialized equipment to determine the distance between wave peaks.

5. What factors can affect the wavelength of light?

The wavelength of light can be affected by the medium through which it travels, such as air, water, or a vacuum. It can also be affected by the source of the light, such as an incandescent bulb, fluorescent light, or laser. Additionally, the wavelength can be altered by passing through filters or by interacting with other objects, such as a prism.

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