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 \times 10^{2} \mathrm{~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 2 t traveled by ray (2). The wavelength is \lambda / n, where n is the index of refraction in the substance.
Use the appropriate condition for destructive interference in a thin film:
2 t=\frac{\lambda}{2 n}
Solve for the thickness t and substitute:
t=\frac{\lambda}{4 n}=\frac{7.80 \times 10^{2} \mathrm{~nm}}{(4)(1.60)}=1.22 \times 10^{2} \mathrm{~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.