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## Q. 11.6

A CD stores 700 MB of data. The region of the CD that stores data has an inner radius of 2.30 cm and outer radius of 5.80 cm, and the width of the data track is 1.60 m. What is the surface area allowed for each data bit? What are the approximate dimensions of each bit along the track?

Strategy Note that 1 byte (B) = 8 bits. The usable surface area is the difference between the area of a disk of radius 5.80 cm and a disk of radius 2.30 cm—effectively subtracting out the “hole” in the middle of the disk. Then the area per bit is that surface area divided by the number of bits

## Verified Solution

The usable area is

$A=\pi\left[(0.058 m )^{2}-(0.023 m )^{2}\right]=8.91 \times 10^{-3} m ^{2}$

The number of bits is $700 MB \times 8 \text { bits } / B =5.60 \times 10^{9} \text { bits }$. The area per bit is

$\frac{\text { Area }}{\text { Bit }}=\frac{8.91 \times 10^{-3} m ^{2}}{5.60 \times 10^{9} bits }=1.6 \times 10^{-12} m ^{2} / bit$

With a track width of 1.6 μm, the length of a bit (area divided by width) is

$\text { Bit length }=\frac{1.6 \times 10^{-12} m ^{2}}{1.6 \times 10^{-6} m }=1.0 \times 10^{-6} m$

The bit length is just 1 micron. Thus the length and width of the bit are comparable to the IR laser wavelength (780 nm) used to scan it.