Red and second light through double slits: wavelength of second light

[vtnsx]

I'm so lost... can someone please explain :( thanx

Red light with wavelength 700 nm is passed through a two-slit apparatus. At the same time, monochromatic visible light with another wvelength passes through the same apparatus. As a result, most of the pattern that appears on the screen is a mixture of two colors; however, the center of the third bright fringe (m=3) of the red light appears pure red, with none of other color. What are the possible wavelengths of the second type of visible light? Do you need to know the slit spacing to answer this question? Why or why not?

 

[learningphysics]

Hmmm... I think I understand the question. Your physics text should have a derivation for the distance of a bright fringe from the central bright fringe.

1) Find the distance (or a formula for the distance) for the 3rd bright fringe of the red light.

2) The other wavelength of light cannot have a fringe at the same distance as what the distance you found in part 1 (since the center of the 3rd fringe is all red). Use this fact to obtain the possible values of the other wavelength. You'll be using the same distance equation. (hint: what wavelengths can the other light NOT have)

From 1) and 2) you should be able to find out whether or not the slit spacing matters.

 

[wais]

Hi there,

I can understand your frustration with this topic, but don't worry, I'll do my best to explain it to you!

First, let's start with the concept of intensity. Intensity is the amount of energy that is being transmitted through a certain area. In physics, it is often represented by the symbol I and is measured in units of watts per square meter (W/m²). In the context of double slit experiments, intensity refers to the brightness of the fringes or interference pattern that is observed on a screen when light passes through two closely spaced parallel slits.

Now, onto the double slit experiment itself. This experiment is a classic demonstration of wave interference and is used to study the behavior of light. In this experiment, a light source, in this case, a red light with a wavelength of 700 nm, is shone through two parallel slits. The light that passes through the slits diffracts and creates an interference pattern on a screen behind the slits.

Now, when you introduce a second light source with a different wavelength, let's say blue light with a wavelength of 450 nm, the interference pattern on the screen changes. This is because the two wavelengths of light have different frequencies and therefore, different interference patterns. The resulting pattern is a mixture of the two colors, with some regions appearing brighter and others appearing darker.

However, there is one specific point on the screen, the third bright fringe (m=3), where the red light appears pure red, with no traces of the blue light. This is because at this point, the two wavelengths are in phase, meaning they have the same frequency and their peaks and troughs align perfectly, resulting in constructive interference and a pure red color.

To answer your question, the possible wavelengths of the second type of visible light can be calculated using the equation for constructive interference: d sinθ = mλ, where d is the distance between the slits, θ is the angle of diffraction, m is the order of the bright fringe, and λ is the wavelength of light. Since we know the values for d, m, and λ for the red light, we can rearrange the equation to solve for the value of λ for the blue light.

In this case, we do not need to know the slit spacing to answer the question, as it is already given to us. However, in other scenarios, where the slit spacing is not known, we would need to know