A compact disc (CD) is read from the bottom by a semiconductor laser with wavelength 790 nm passing through a plastic substrate of refractive index 1.8. When the beam encounters a pit, part of the beam is reflected from the pit and part from the flat region between the pits, so these two beams interfere with each other (Fig. E35.31). What must the minimum pit depth be so that the part of the beam reflected from a pit cancels the part of the beam reflected from the flat region? (It is this cancellation that allows the player to recognize the beginning and end of a pit.

We are given${\lambda }_{air}=790nm\&n_{plastic}=1.8$ . In this case, both reflected from the same surface of the plastic material. This means that whether both will experience a phase change or they will experience no phase change at all.

Hence, we need the two reflected rays are in same phase.

We need the two reflected rays to interact destructively, this means that the net distance between the two tips will be given by

$2t=\left(m+\frac{1}{2}\right){\lambda }_{plastic}$

Nothing that laser beam reflected after traveling through the plastic substance.

So, the height t between the two tips is given by

$t=\left(m+\frac{1}{2}\right)\frac{{\lambda }_{plastic}}{2}$ (1)

As from Snell’s Law

$$\begin{gathered} n_1{\lambda }_1=n_2{\lambda }_2 \\ {n}_{air}{\lambda }_{air}=n_{plastic}{\lambda }_{plastic} \end{gathered}$$

Now solve for${\lambda }_{plastic}$

$$\begin{gathered} {\lambda }_{plastic}=\frac{{\lambda }_{air}}{n_{plastic}} \\ t=\left(m+\frac{1}{2}\right)\frac{{\lambda }_{air}}{2n_{plastic}} \end{gathered}$$

For the minimum pit depth, m =0

$t=\left(0+\frac{1}{2}\right)\frac{{\lambda }_{air}}{2n_{plastic}}$

$$\begin{gathered} t=\frac{{\lambda }_{air}}{4n_{plastic}} \\ t=\frac{790}{4\times 1.8} \\ t=110nm \end{gathered}$$

Hence, the minimum pit depth should be 110 nm .