Question 24.5: Pit Depth in a CD GOAL Apply interference principles to a CD...
Pit Depth in a CD
GOAL Apply interference principles to a CD.
PROBLEM Find the pit depth in a CD that has a plastic transparent layer with index of refraction of 1.60 and is designed for use in a CD player using a laser with a wavelength of 7.80 × 10² nm in air.
STRATEGY (See Fig. 24.12.) Rays 1 and 2 both reflect from the metal layer, which acts like a mirror, so there is no phase difference due to reflection between those rays. There is, however, the usual phase difference caused by the extra distance 2t traveled by ray 2. The wavelength is λ/n, where n is the index of refraction in the substance.
Use the appropriate condition for destructive interference in a thin film:
2t={\frac{\lambda}{2n}}
Solve for the thickness t and substitute:
t={\frac{\lambda}{4n}}={\frac{7.80~\times~10^{2}\;{\mathrm{nm}}}{(4)(1.60)}}= 1.22 × 10² nm
REMARKS Different CD systems have different tolerances for scratches. Anything that changes the reflective properties of the disk can affect the readability of the disk.