Find the height \(h\) of the pits on a CD (Fig. \(25.6 \mathrm{a}\) ). When the laser beam reflects partly from a pit and partly from land (the flat aluminum surface) on either side of the "pit," the two reflected beams interfere destructively; \(h\) is chosen to be the smallest possible height that causes destructive interference. The wavelength of the laser is \(780 \mathrm{nm}\) and the index of refraction of the poly carbonate plastic is \(n=1.55.\)

Optical path length (OPL) is the product of the index of refraction and the physical distance traveled by a light wave. In this scenario, the light reflecting from the pit and the land will have different optical paths.

When the light reflects partly from the pit and from the land, the two beams will interfere destructively which means their optical path length difference (\(\Delta\)OPL) should be an odd multiple of the half-wavelength. Mathematically, this can be represented as: \(\Delta \text{OPL} = (2k + 1) \frac{\lambda}{2}\) where \(k\) is an integer, and \(\lambda\) is the wavelength of the light in vacuum.

Since the light is passing through polycarbonate plastic, we need to take into account the index of refraction (given as \(n = 1.55\)). We can find the wavelength of the light in the plastic medium by dividing the vacuum wavelength by the index of refraction: \(\lambda_{\text{plastic}} = \frac{\lambda}{n} = \frac{780 \text{ nm}}{1.55}\)

Now, we should find the smallest height \(h\) that will cause destructive interference. Since the light enters and exits the pit, the actual physical distance traveled by the light in the pit is \(2h\). So, the optical path length difference is \(\Delta\)OPL \(= 2n h - 2h\), where \(n h\) is the OPL for pit region and \(2h\) is the OPL for land region. We can plug the expression for the \(\Delta\)OPL from Step 2 into the equation: \((2k + 1) \frac{\lambda}{2} = 2n h - 2h\). Rearrange and substitute the known values to find the height of the pit: \(h=\frac{(2k+1)\frac{\lambda_{\text{plastic}}}{2}}{2n-2}\) To find the minimum height, let \(k = 0\). This gives: \(h = \frac{\frac{\lambda_{\text{plastic}}}{2}}{2n-2}=\frac{\frac{780\,\text{nm}/1.55}{2}}{2\times 1.55-2}\)

Solve the equation to find the height of the pit: \(h \approx 126 \,\text{nm}\) The height of the pits on the CD is approximately \(126\,\text{nm}\).