Possible Wavelength Difference Between Two Rays Passing Through a Prism?

[Stacey Freels]

Homework Statement

The two rays shown below, a and b, have different wavelengths. They travel through the glass prism as shown; 1) is this possible? 2) If this is possible, which has the longer wavelength, ray a or ray b?

[see attached figure]

Homework Equations

λ = λi/n
critical angle = arcsin(n2/n1)

The Attempt at a Solution

We are trying to think through this but we keep hitting mental blocks. We know that a and b have different wavelengths, and that both of the wavelengths will change by the same factor when they enter the glass. We also know that you will get total internal reflection (TIR) at some angle specified by the indices of refraction for the glass prism and air.

We aren't totally convinced that this is possible, because the critical angle depends wholly on the indices of refraction, which are independent of wavelength. Is this correct reasoning? By this logic, it is not possible.

Thank you so much!

 

[haruspex]

indices of refraction, which are independent of wavelength.

If that is the case, how does a prism separate colours?

 

[Stacey Freels]

If that is the case, how does a prism separate colours?

True. How can we relate wavelength and indices of refraction?

 

[haruspex]

True. How can we relate wavelength and indices of refraction?

Look up dispersion.

 

[Stacey Freels]

Look up dispersion.

If we use the refractive indices for red light and purple light, for example, we get 1.54 for red light and 1.59 for purple light (given in our textbook). In the diagram we see that one ray (ray A) is internally reflecting, while ray B is still refracting. By the diagram, it appears that the two rays are hitting the side of the prism at 45 degrees. If this is true, then both ray A and ray B should be internally reflecting because 45 degrees is greater than the critical angle (40.49 for 1.54 and 38.97 for 1.59). Are there two wavelengths that would make the difference in n great enough that ray A would internally reflect and ray B would refract?

 

[haruspex]

it appears that the two rays are hitting the side of the prism at 45 degrees.

Not according to the image you posted.

 

[Stacey Freels]

Not according to the image you posted.

What if we took the rays out of the visible light spectrum? Does light at these wavelengths behave how visible light does when it hits a prism?

 

[haruspex]

What if we took the rays out of the visible light spectrum? Does light at these wavelengths behave how visible light does when it hits a prism?

Glass is not necessarily transparent at other wavelengths.
But why do you want to? Did you understand my previous reply?

 

[Stacey Freels]

Glass is not necessarily transparent at other wavelengths.
But why do you want to? Did you understand my previous reply?

Just want to make sure that this scenario is actually impossible. If the wavelengths were out of the visible spectrum (say 1000nm and 200nm), the index of refraction for these wavelengths would differ even more and there could be wavelengths where ray A internally reflects and ray B refracts. In this example, if we use red and purple light, the angle would have to be ~39 degrees as it hits the edge of the prism to give this scenario. If the diagram is correct, and we are limited to the visible light spectrum, then yes this scenario is impossible. If the angle is not exactly 45 degrees, or we are not limited to the visible light spectrum, then the scenario could still be possible.

 

[haruspex]

if we use red and purple light, the angle would have to be ~39 degrees

Why are you so sure it isn't?

 

[Stacey Freels]

Why are you so sure it isn't?

Because in the diagram it appears to be ~45 degrees

 

[Stacey Freels]

Why are you so sure it isn't?

But anyway, it could be the diagram is not perfect. So, if we assume that the angle is 39 degrees, ray A would have a shorter wavelength than ray B, correct?

 

[haruspex]

Because in the diagram it appears to be ~45 degrees

As I wrote, in the diagram you posted, it looks significantly less than 45 degrees. I measured the triangle sides and calculated the incidence angle as 40.9 degrees.

 

[haruspex]

But is it important? There seems to be only one refraction.

I (we) believe the diagram intends to show two rays of different wavelength entering on the same path, one undergoing total internal reflection, the other emerging, refracted, from the hypotenuse. The OP is trying to understand whether this can be true for visible light in glass, and has determined it requires the incidence angle to be around 40 degrees. I maintain it is about that in the diagram.

 

[fresh_42]

I (we) believe the diagram intends to show two rays of different wavelength entering on the same path, one undergoing total internal reflection, the other emerging, refracted, from the hypotenuse. The OP is trying to understand whether this can be true for visible light in glass, and has determined it requires the incidence angle to be around 40 degrees. I maintain it is about that in the diagram.

I thought it should have been one ray entering and two leaving. But even if there were two rays, how can we draw any conclusions about the wavelengths, for there is only one refraction. Or do we assume there are actually two refractions at the splitting point? One by 45° looking like a reflection?
(I don't know anything better here, just try to understand.)

 

[haruspex]

I thought it should have been one ray entering and two leaving.

It says A and B, shown leaving, have different wavelengths. Unless the prism is magically changing the frequencies, that means two at different wavelengths entered.

how can we draw any conclusions about the wavelength

The refractive index depends on the wavelength. If the two rays are both visible light then that limits the range of indices. The question is whether the diagram can be reasonably interpreted as showing an angle of incidence that would allow one wavelength of visible light to be internally reflected but not the other.

 

[fresh_42]

I thought this could even happen with one monochromatic ray.

 

[haruspex]

I thought this could even happen with one monochromatic ray.

Strictly speaking, yes. At incidence less than the critical angle, there is some reflection as well as refraction. But the question clearly says the two emergent rays are at different wavelengths. There should be a third ray, the partial reflection of ray B, but the omission of that is not important.

 

[haruspex]

As I wrote, in the diagram you posted, it looks significantly less than 45 degrees. I measured the triangle sides and calculated the incidence angle as 40.9 degrees.

I just realized what is wrong with the diagram, and this may be causing your problem. The reflection shown is at the wrong angle. It should be the same as the incident angle, so should slope down to the left, not be horizontal.

 

[ehild]

If the angle of incidence is not 45° the ray would change direction when leaving the prism on the other side. We can assume 45°as incident angle. But the prism can be in some medium different from air. And there are glasses with quite high refractive indices.

 

[Stacey Freels]

If the angle of incidence is not 45° the ray would change direction when leaving the prism on the other side. We can assume 45°as incident angle. But the prism can be in some medium different from air. And there are glasses with quite high refractive indices.

Isn't the ray changing direction? And why would an angle of 45° result in the ray not changing direction? Also, I'm having a bit of trouble connecting what you are saying to the problem itself. Could you please clarify? Thank you!

 

[ehild]

Isn't the ray changing direction? And why would an angle of 45° result in the ray not changing direction? Also, I'm having a bit of trouble connecting what you are saying to the problem itself. Could you please clarify? Thank you!

See figure. If it is an equilateral right triangle, and the light hits the side a perpendicularly, it is incident on side b at 45°, reflected at 45°and incident perpendicularly at side c. At normal incidence, the transmitted light does not change direction.

 

[haruspex]

We can assume 45°as incident angle

I disagree. The slope of the hypotenuse is clearly not 45 degrees. Indeed, it is about 40 degrees, as it needs to be to make it feasible that two visible rays will behave differently. I think it more likely that the reflected ray was drawn carelessly. The reflection angle is clearly larger than the incidence angle.