Double Slit Problem: Solve for D2-D1 with 500nm Wavelength

[knox_122]

here the problem:

In the Young Double Slit experiment, alternatingbands of bright and dark regions are produced on the screen. At the dark band shown in the picture below, on slit is a closer than the other slit: In other words, D1 is less than D2, Find D2-D1, Assuming that the light has a wavelength of 500nm.


The Picture he shows, he said is not to scale so it won't help. there are no other measurements listed. How do i go about figuering this out?

 

[lightgrav]

I hope there's an angle scale in the pic.

Distance between slit centers "d" makes fringes (bands) that are angle-dependent, as $$m \lambda = d sin \theta$$

 

[quicknote]

The double slit problem is a classic demonstration of the wave-like nature of light. In order to solve for D2-D1, we need to use the equation for the position of the dark fringes:

dsinθ = mλ

Where d is the distance between the two slits, θ is the angle between the center of the screen and the position of the dark fringe, m is the order of the dark fringe (1, 2, 3, etc.), and λ is the wavelength of the light.

Since we are given that the light has a wavelength of 500nm (or 0.5μm), we can plug this value into the equation and solve for D2-D1. However, as the picture provided is not to scale and there are no other measurements listed, we cannot accurately determine the values for d and θ. Therefore, we cannot solve for D2-D1 without more information.

To accurately solve for D2-D1, we would need to know the distance between the slits (d) and the angle (θ) at which the dark fringe is observed. Without these values, it is impossible to solve for D2-D1.

In conclusion, the information provided is not sufficient to solve for D2-D1 in the double slit problem. More measurements and accurate scaling are necessary to accurately solve for this value.