Question Video: Calculating the Speed of Light in a Different Material | Nagwa Question Video: Calculating the Speed of Light in a Different Material | Nagwa

Question Video: Calculating the Speed of Light in a Different Material Physics • Second Year of Secondary School

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The diagram shows the wave fronts of a light wave passing from air into a different material. The light wave moves at 3 × 10⁸ m/s in air. At what speed does the light wave move in the other material? Give your answer to three significant figures.

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Video Transcript

The diagram shows the wave fronts of a light wave passing from air into a different material. The light wave moves at three times 10 to the eight meters per second in air. At what speed does the light wave move in the other material? Give your answer to three significant figures.

To begin, we should remember that we can use wave fronts, which look like parallel lines, to represent subsequent points in the cycle of a transverse wave. The wave itself travels in a direction perpendicular to the wave fronts. Here, the wave is moving from left to right. And so the distances here, given in nanometers, show us the wavelengths of this light wave as it travels from air to some other medium. Note that as the wave enters this different medium, its wavelength changes. We should recall that this corresponds to the speed of the wave changing as it passes into the new medium, and it’s our job to find the wave’s new speed. So the speed and wavelength change. But recall that as a wave enters a new medium, its frequency must remain the same.

At this point, it’ll be helpful to recall an equation that relates the speed of a wave, 𝑣, to its frequency and wavelength. 𝑣 equals 𝑓 times 𝜆, where 𝑓 is the frequency and 𝜆 is the wavelength. We know that in air, the speed of a light wave equals three times 10 to the eight meters per second. In this question, the light wave begins traveling in air. So this is the speed of the wave, 𝑣, before it passes into the new medium. To stay organized, we can use the subscript one to denote properties of the wave while it’s traveling in air. So let’s also call its frequency here 𝑓 one and its wavelength 𝜆 one.

Applying these terms to the equation for the speed of a wave, we have 𝑣 one equals 𝑓 one times 𝜆 one. Similarly, in the new medium, the wave will have a frequency 𝑓 two, a wavelength of 𝜆 two, and a speed of 𝑣 two. So we now have another equation, 𝑣 two equals 𝑓 two times 𝜆 two. Remember that we’ve already established that the frequency of the wave in either material is the same. So we can say that 𝑓 one equals 𝑓 two. And to simplify the math, we can just use a generic 𝑓 to represent the frequency. So let’s make that substitution in our equations.

Let’s also take a moment to write out all the known values here. From the diagram, we can see that 600 nanometers is the wavelength of the light before it enters the new medium, 𝜆 one. And in the new medium its wavelength is 350 nanometers. This is 𝜆 two. So we’ve been given the speed of light in air and the two different wavelengths of the light wave and need to use this information to find its speed in the new medium. Let’s do a little bit of algebra to find a way to solve for 𝑣 two expressed in terms of our known values.

Notice that both of our two equations share a similar term, 𝑓. We’ll use this as a way to relate the equations to each other. So let’s rearrange equation one to make frequency the subject. All we need to do is divide both sides by 𝜆 one. So that term cancels out of the right-hand side, leaving 𝑓 by itself. Thus, the expression can be written as 𝑓 equals 𝑣 one over 𝜆 one. Now we can substitute this in for the frequency in equation two. So we have 𝑣 two equals 𝑣 one over 𝜆 one times 𝜆 two. Thus, we have an expression for 𝑣 two in terms of our known values. So let’s now substitute values for the terms on the right-hand side.

Before we calculate though, we need to remember to convert nanometers into meters by recalling that one nanometer equals 10 to the negative nine meters. And still looking at the units here, notice that we have meters per second divided by meters and then multiplied by meters. The meters terms in the numerator and denominator cancel each other out, leaving just meters per second, which is a good sign since we are solving for a speed after all.

Finally, plugging this into a calculator, we get a result of 1.75 times 10 to the eight meters per second to three significant figures. This then is our final answer. We’ve found that in this other material, the light wave travels at a speed of 1.75 times 10 to the eight meters per second.

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