Video: Determining the Paths Traveled by Dispersed Light Rays

A narrow beam of light contains light of wavelength πœ† = 580 nm and of wavelength πœ† = 550 nm. The light passes from polystyrene to air at angle of exactly 30Β° to the polystyrene-air boundary. The refractive index of the polystyrene for 550 nm wavelength is 1.4930 and the refractive index of the polystyrene is 1.4920 for 580 nm wavelength light. Use 1.00029 as the refractive index of air for both wavelengths. What angle separates the directions of propagation of the 580 nm wavelength and the 550 nm wavelength parts of the beam as they enter air? The 580 nm wavelength and 550 nm wavelength parts of the emergent light beam are incident on a surface that is perpendicular to the axis of propagation of the emergent beam. At the surface, the 580 nm wavelength and 550 nm wavelength parts of the beam are separated from each other by 1.00 mm. What is the distance between the point of emergence of the beam and this surface?

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Video Transcript

A narrow beam of light contains light of wavelength πœ† equals 580 nanometers and of wavelength πœ† equals 550 nanometers. The light passes from polystyrene to air at angle of exactly 30 degrees to the polystyrene-air boundary. The refractive index of the polystyrene for 550 nanometer wavelength is 1.4930 and the refractive index of the polystyrene is 1.4920 for 580 nanometer wavelength light. Use 1.00029 as the refractive index of air for both wavelengths. What angle separates the directions of propagation of the 580 nanometer wavelength and the 550 nanometer wavelength parts of the beam as they enter air? The 580 nanometer wavelength and 550 nanometer wavelength parts of the emergent light beam are incident on a surface that is perpendicular to the axis of propagation of the emergent beam. At the surface, the 580 nanometer wavelength and 550 nanometer wavelength parts of the beam are separated from each other by 1.00 millimeters. What is the distance between the point of emergence of the beam and this surface?

In this problem statement, we’re told several helpful facts. We’re told that one of the wavelengths of light, that we’ll call πœ† sub one, is equal to 580 nanometers. And we’re also told that another wavelength which we’ll call πœ† sub two equals 550 nanometers. We’re told that light that includes these two wavelengths travels toward an interface with polystyrene on top and air on bottom. The ray incident on the interface comes in at an angle of 30 degrees. And we’ll call that angle πœƒ sub 𝑖. When the incident ray reaches the interface, these two wavelengths, πœ† one and πœ† two, are refracted differently depending on their wavelength. And the difference between the angles of refraction of these two wavelengths, we can call Ξ”πœƒ. That’s the first thing we’re trying to solve for.

In addition to this information, we’re told that the index of refraction of the polystyrene varies with πœ†. When πœ† equals 580 nanometers, the index of refraction of the polystyrene is 1.4920. And when πœ† equals 550 nanometers, the polystyrene index of refraction is 1.4930. We’re also told that for both these wavelengths, we can treat the index of refraction of air as a constant, 1.00029. To move towards solving for Ξ”πœƒ, the difference in the refracted angles between these two wavelengths, let’s recall Snell’s law. This law says that if we take the product of the index of refraction of the material through which a ray is incident on an interface with the sine of the angle of incidence, then that’s equal to π‘›π‘Ÿ, the index of refraction of the second material, times sine of the refracted angle, πœƒ sub π‘Ÿ. When we apply Snell’s law to our situation, we write that the index of refraction of polystyrene times the sine of the incident angle is equal to the index of refraction of air times the sine of the refracted angle.

In our case, we have two values for 𝑛 sub 𝑝 and as a result, two values for πœƒ sub π‘Ÿ. So we can write out two separate versions of Snell’s law. In the first version, we write the 𝑛 of polystyrene when πœ† is 580 nanometers times the sine of πœƒ sub 𝑖 is equal to the index of refraction of air times the sine of the refracted angle when πœ† is 580 nanometers. The second version is identical to the first one except that every time we see 580 nanometers, we replace it with 550. Now we’ll rearrange some of our given information as it’s laid out on the screen to create space to plug in the numbers for these two equations.

We can now rearrange this first expression of Snell’s law to solve for the value we’re interested in, which is the refracted angle when πœ† is 580 nanometers. When we do that, we find that πœƒ, the refracted angle, when πœ† is 580 nanometers is the arc sine of the index of refraction of polystyrene at that wavelength times the sine of the incident angle divided by the index of refraction of air. Likewise, when we rearrange to solve for πœƒ sub 550, we find a similar expression. Now Ξ”πœƒ, which is the difference between these two angles of refraction, is equal to the magnitude of πœƒ sub 580 minus πœƒ sub 550.

When we plug-in for the values of the index of refraction of polystyrene; πœƒ sub 𝑖, the incident angle of the beam; and the index of refraction of air and enter these values into our calculator, we find that the magnitude of the difference of these refraction angles, Ξ”πœƒ, is equal to 0.043 degrees. That’s the difference between the refracted angle of the light at one wavelength and the light at the other wavelength. So that’s part one of our problem.

In part two, we imagine that if we take our diagram and extend out the refracted rays onto a screen some distance away from the interface between polystyrene and air, then the two rays of light, one at one wavelength πœ† one and one at the other, strike the screen separated by a distance of 1.00 millimeters. In part two of this problem, we want to find out what is 𝑑. That is, what is the distance between this screen and the interface where the beam of light splits.

To solve for 𝑑, let’s first draw the diagram in a slightly different orientation. If we redraw the split rays so that the screen that they hit is vertically aligned, then we see that the angle between these rays β€” which we solved for in part one, Ξ”πœƒ β€” can be bisected by a line that creates two right triangles. If we look at the top one of these triangles, then we know that the angle on the left corner is Ξ”πœƒ divided by two. The opposite side of the triangle is one-half a millimeter. And we know that the adjacent leg is a length 𝑑.

Focusing on this triangle, we can write that the tangent of the angle Ξ”πœƒ divided by two is equal to 0.50 millimeters divided by 𝑑. In other words, 𝑑 is equal to 0.50 millimeters divided by the tangent of Ξ”πœƒ over two. When we plug-in the value for Ξ”πœƒ found earlier and calculate this value, we find that, in meters to two significant figures, 𝑑 equals 1.3 meters. That’s the distance the screen must be from the interface where the beam of light splits.

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