Depth of pit of CD - why 1/4 of wavelength?

[huey910]

Homework Statement

Why does the depth of the pit have to be 1/4 of the wavelength?

Homework Equations

I understand that this causes two beams to interfere destructively because the path difference resulted is 1/2 of a wavelength - however, I cannot prove this geometrically - can somebody do it?

The Attempt at a Solution

there must be a special geometric construction of the two beams that interfere destructively, how does one know that this happens everytime two beams are incident on the CD? Also, I do not understand how the depth of the pit can geometrically imply that the path difference between two beams that are incident on the land and the pit is 1/2 of the wavelength. Please advise, thank you

 

[anathema]

.

Dear forum post author,

Thank you for your question. I would like to address your concerns and provide a geometric explanation for why the depth of the pit must be 1/4 of the wavelength for destructive interference to occur.

Firstly, it is important to understand that when light waves interact, they exhibit properties of both particles and waves. This means that they can interfere with each other, just like waves in water. When two light waves intersect, they can either reinforce each other (constructive interference) or cancel each other out (destructive interference).

In the case of a CD, the surface of the CD acts as a diffraction grating, which means it can split a beam of light into multiple beams. When these beams reflect off the surface of the CD, they interfere with each other. The depth of the pit on the CD is crucial because it determines the path difference between the two beams that are incident on the land and the pit.

To understand this, let's imagine two beams of light, A and B, that are incident on the CD surface. Beam A hits the land on the CD surface and reflects off it, while beam B hits the pit and reflects off it. When these two beams meet, they will interfere with each other. However, for destructive interference to occur, the path difference between the two beams must be 1/2 of the wavelength.

Now, let's consider the depth of the pit. As you correctly mentioned, when the depth of the pit is 1/4 of the wavelength, the path difference between beams A and B will be 1/2 of the wavelength. This is because beam A travels a distance of 1/2 of the wavelength (due to reflection at the land) while beam B travels a distance of 1/4 of the wavelength (due to reflection at the pit). This results in a path difference of 1/2 of the wavelength, leading to destructive interference.

In conclusion, the depth of the pit on a CD must be 1/4 of the wavelength for destructive interference to occur. This is because it creates a path difference of 1/2 of the wavelength between the two interfering beams. I hope this explanation helps. If you have any further questions, please do not hesitate to ask.
Scientist