An advanced computer sends information to its various parts via infrared light pulses traveling through silicon fibers. To acquire data from memory, the central processing unit sends a light-pulse request to the memory unit. The memory unit processes the request, then sends a data pulse back to the central processing unit. The memory unit takes $0.5ns$ to process a request. If the information has to be obtained from memory in $2.0ns$, what is the maximum distance the memory unit can be from the central processing unit?

We have been given that an advanced computer sends information to its various parts via infrared light pulses traveling through silicon fibers. The memory unit takes $0.5ns$to process a request. The time taken by the memory unit to deliver the information back to the central processing unit is $2.0ns$.

We need to find the the maximum distance between the memory unit and the central processing unit.

The refractive index of silicon is $n=3.50$.

The speed of light through the silicon fibers can be calculated by :

$v=\frac{c}{n}$, where $v$is speed of light in silicon fibers, $c$is speed of light in vacuum, $n$is the refractive index of silicon.

Substituting $c=3.0\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$and $n=3.50$, we get:

localid="1650362408425" $v=\frac{3.0\times {10}^8{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}}{3.50}$

On simplifying, we get:

$v=0.857\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

The time taken by the infrared light to travel to the memory unit from the central processing unit is

$$\begin{gathered} ∆t=(2.0-0.5)ns \\ ∆t=1.5ns \end{gathered}$$

We can find the total distance travelled by the infrared light by the following formula:

$d_{total}=v∆t$

Substituting the values of $v$and $∆t$, we get:

$$\begin{gathered} d_{total}=(0.857\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.)(1.5ns)\left(\frac{1s}{{10}^{9}ns}\right) \\ {d}_{total}=\frac{9}{70}m \end{gathered}$$

The distance between the central processing unit and the memory unit is half of the total distance travelled:

$d=\frac{d_{total}}{2}$

Substituting the value of $d_{total}$, we get:

$d=\frac{9m}{2\left(70\right)}$

$d=6.4\times {10}^{-2}m$

$$\begin{gathered} d=(6.4\times {10}^{-2}m)\left(\frac{100cm}{1m}\right) \\ d=6.4cm \end{gathered}$$