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

Question

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

Accepted Answer

The usable area is

[latex]A=\pi\left[(0.058 m )^{2}-(0.023 m )^{2} ight]=8.91 imes 10^{-3} m ^{2}[/latex]

The number of bits is [latex]700 MB imes 8 ext { bits } / B =5.60 imes 10^{9} ext { bits }[/latex]. The area per bit is

[latex]\frac{ ext { Area }}{ ext { Bit }}=\frac{8.91 imes 10^{-3} m ^{2}}{5.60 imes 10^{9} bits }=1.6 imes 10^{-12} m ^{2} / bit[/latex]

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

[latex] ext { Bit length }=\frac{1.6 imes 10^{-12} m ^{2}}{1.6 imes 10^{-6} m }=1.0 imes 10^{-6} m[/latex]

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.

Question 11.6: A CD stores 700 MB of data. The region of the CD that stores...

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