91Ó°ÊÓ

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

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

On simplifying, we get:

$v=0.857×{10}^8\raisebox{1ex}{$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:

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

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:

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

$d=6.4×{10}^{-2}m$

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