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QuestionCalculate the unique spots a laser with diameter 4.05μm4.05 \mu \mathrm{m} can read on a DVD with a 4584 \frac{5}{8} inch diameter.

Studdy Solution

STEP 1

Assumptions1. The diameter of the DVD is 4584 \frac{5}{8} inches. . The diameter of the center spindle region that cannot be written is 1381 \frac{3}{8} inches.
3. The diameter of the laser beam is 4.05μm4.05 \mu m.
4. The area of a circle is A=πrA=\pi r^{}.
5. The number of unique spots on the disk that the laser could read is the ratio of the writable area of the disk to the area of the laser beam.

STEP 2

First, we need to convert the diameters of the DVD and the center spindle region from mixed numbers to improper fractions.
iameterDVD=458=378iameter_{DVD} =4 \frac{5}{8} = \frac{37}{8}iameterspindle=18=118iameter_{spindle} =1 \frac{}{8} = \frac{11}{8}

STEP 3

Now, we convert the diameters from inches to meters using the conversion factors.
iameterDVD=378×2.54×10−2iameter_{DVD} = \frac{37}{8} \times2.54 \times10^{-2}iameterspindle=118×2.54×10−2iameter_{spindle} = \frac{11}{8} \times2.54 \times10^{-2}

STEP 4

Calculate the diameters in meters.
iameterDVD=378×2.54×10−2=0.117475iameter_{DVD} = \frac{37}{8} \times2.54 \times10^{-2} =0.117475iameterspindle=118×2.54×10−2=0.034925iameter_{spindle} = \frac{11}{8} \times2.54 \times10^{-2} =0.034925

STEP 5

Now, we calculate the radii of the DVD and the center spindle region by dividing the diameters by2.
RadiusDVD=iameterDVD2Radius_{DVD} = \frac{iameter_{DVD}}{2}Radiusspindle=iameterspindle2Radius_{spindle} = \frac{iameter_{spindle}}{2}

STEP 6

Calculate the radii in meters.
RadiusDVD=0.1174752=0.0587375Radius_{DVD} = \frac{0.117475}{2} =0.0587375Radiusspindle=0.0349252=0.0174625Radius_{spindle} = \frac{0.034925}{2} =0.0174625

STEP 7

Now, we calculate the writable area of the DVD by subtracting the area of the center spindle region from the total area of the DVD.
Areawritable=π(RadiusDVD)2−π(Radiusspindle)2Area_{writable} = \pi (Radius_{DVD})^{2} - \pi (Radius_{spindle})^{2}

STEP 8

Plug in the values for the radii to calculate the writable area.
Areawritable=π(0.0587375)2−π(0.0174625)2Area_{writable} = \pi (0.0587375)^{2} - \pi (0.0174625)^{2}

STEP 9

Calculate the writable area in square meters.
Areawritable=π(.0587375)2−π(.0174625)2=.00292147Area_{writable} = \pi (.0587375)^{2} - \pi (.0174625)^{2} =.00292147

STEP 10

Now, we calculate the area of the laser beam. First, we convert the diameter of the laser beam from micrometers to meters.
iameterlaser=4.05×10−6iameter_{laser} =4.05 \times10^{-6}

STEP 11

Calculate the radius of the laser beam by dividing the diameter by.
Radiuslaser=iameterlaserRadius_{laser} = \frac{iameter_{laser}}{}

STEP 12

Calculate the radius in meters.
Radiuslaser=4.05×10−62=2.025×10−6Radius_{laser} = \frac{4.05 \times10^{-6}}{2} =2.025 \times10^{-6}

STEP 13

Calculate the area of the laser beam.
Arealaser=π(Radiuslaser)2Area_{laser} = \pi (Radius_{laser})^{2}

STEP 14

Plug in the value for the radius to calculate the area of the laser beam.
Arealaser=π(2.025×10−6)2Area_{laser} = \pi (2.025 \times10^{-6})^{2}

STEP 15

Calculate the area of the laser beam in square meters.
Arealaser=π(2.025×10−)2=.29445×10−11Area_{laser} = \pi (2.025 \times10^{-})^{2} =.29445 \times10^{-11}

STEP 16

Now that we have the writable area of the DVD and the area of the laser beam, we can calculate the number of unique spots on the disk that the laser could read. This is done by dividing the writable area by the area of the laser beam.
Number of spots=AreawritableArealaserNumber\, of\, spots = \frac{Area_{writable}}{Area_{laser}}

STEP 17

Plug in the values for the writable area and the area of the laser beam to calculate the number of spots.
Number of spots=0.00292147.29445×10−11Number\, of\, spots = \frac{0.00292147}{.29445 \times10^{-11}}

STEP 18

Calculate the number of unique spots on the disk that the laser could read.
Number of spots=0.00292147.29445×10−11=2.256×1014Number\, of\, spots = \frac{0.00292147}{.29445 \times10^{-11}} =2.256 \times10^{14}The laser could read approximately 2.256×10142.256 \times10^{14} unique spots on the disk.

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