# Video: Refractive Index Relationship to Speed of Light Propagation

Components of some computers communicate with each other through optical fibers having a refractive index of 1.55. What time in nanoseconds is required for a signal to travel 0.200 m through such a fiber?

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### Video Transcript

Components of some computers communicate with each other through optical fibers having a refractive index of 1.55. What time in nanoseconds is required for a signal to travel 0.200 meters through such a fiber?

We will call the refractive index of 1.55 given in this statement 𝑛. We’re told that the distance the signal needs to travel is 0.200 meters, which we’ll refer to as 𝑑. We want to solve for a time 𝑡 needed for a signal to travel this distance 𝑑. To begin our solution, let’s first recall the relationship for the index of refraction, 𝑛.

A material’s index of refraction is defined as the speed of light in vacuum, 𝑐, divided by the speed of light through that material, 𝑣. In this exercise, we’ll assume that 𝑐, the speed of light in vacuum, is exactly 3.00 times 10 to the eighth meters per second.

When we apply this relationship to our scenario and recall that the average speed of an object is given by the distance it travels divided by the time it takes to travel that distance, we see we can write our equation for 𝑛 as 𝑐 divided by 𝑑 over 𝑡, which is equal to 𝑐 times 𝑡 over 𝑑.

Rearranging this equation to solve for 𝑡, we see that it’s equal to the index of refraction times the distance divided by the speed of light in vacuum. We know each one of these values and can plug in for them now. When we enter these values on our calculator, we find that 𝑡 is equal to 1.03 times 10 to the negative nine seconds, or 1.03 nanoseconds.

That’s how long it would take a light signal to travel 0.200 meters through a fiber with index of refraction 1.55.