Light Refraction: Bending of light

By: Elisa Mercando

The Main Idea

Light refraction is the wavelengths of light entering different materials, where the speed is different. The refraction of light when it passes through a fast medium to a slow medium causing the light ray to bend between the two mediums.

The wavelength, [math]\displaystyle{ \gamma }[/math], of the radiation can either be shorter or longer (depending on the material). The frequency and period of the oscillation do not change, so the speed of the ray changes ([math]\displaystyle{ speed = \frac{\gamma}{T} }[/math]).

A Mathematical Model

The refraction of light can be explained by Snell's law. Snell's law describes a relationship between the bending of the light rays at the medium and speed of the ray in the two medium.

The rays can be represented as straight lines, and the direction of the incoming and outgoing rays can be expressed as angles between the ray and the normal vector (perpendicular to the surface). We can assign [math]\displaystyle{ \theta_1 }[/math] for the incoming angle of the ray and [math]\displaystyle{ \theta_2 }[/math] for the outgoing angle of the ray. The speed of the ray in the air is [math]\displaystyle{ v_1 }[/math] and the speed in the material is [math]\displaystyle{ v_2 }[/math]. In the time T, the ray travels [math]\displaystyle{ v_1*T }[/math] in the air and [math]\displaystyle{ v_2*T }[/math] in the material. The two triangles formed by the angles share a side, d.

The incoming angle is expressed as [math]\displaystyle{ \sin(\theta_1) = \frac{v_1*T}{d} }[/math] and the outgoing angle is expressed as [math]\displaystyle{ \sin(\theta_2) = \frac{v_2*T}{d} }[/math]. [math]\displaystyle{ \frac{T}{d} }[/math] is the same for both triangles. So, [math]\displaystyle{ \frac{\sin(\theta_1}{v_1} = \frac{\sin(\theta_2}{v_2} }[/math].

The index of refraction is expressed as [math]\displaystyle{ n = c/v }[/math].

Therefore, [math]\displaystyle{ \frac{\sin(\theta_1}{c/n_1} = \frac{\sin(\theta_2}{c/n_2} }[/math]. Then we cancel the common factor, c.

Snell's Law: [math]\displaystyle{ n_1\sin(\theta_1) = n_2\sin(\theta_2) }[/math]

A Computational Model

How do we visualize or predict using this topic. Consider embedding some vpython code here Teach hands-on with GlowScript

Examples

Ray Travels from Water into Glass

A beam of light travels through water and hits a glass surface. The angle [math]\displaystyle{ \theta_1 }[/math] between the incident beam and the normal to the glass surface is 23 degrees. The index of refraction of water is 1.33 and the index of refraction of this type of glass is 1.65. What is the angle [math]\displaystyle{ \theta_2 }[/math]?

Solution:

[math]\displaystyle{ n_1\sin(\theta_1) = n_2\sin(\theta_2) }[/math]

[math]\displaystyle{ \sin(\theta_2) = \frac{n_1\sin(\theta_1}{n_2} }[/math]

[math]\displaystyle{ \theta_2 = arcsin\frac{1.33*\sin(23)}{1.65} = }[/math]18.4 degrees

Total Internal Reflection

We've only discussed light that passes from air to glass causing a decrease in the speed of the ray associated with an increase in the index of refraction.

But what happens if we go in the other direction? Going from a higher index of refraction to a lower index of refraction?

Suppose light goes from water (refraction index=1.33) into air (refraction index=1.0). Consider a ray in water at an angle of 60 degrees to the normal. Now predict the angle the ray will make to the normal in the air.

[math]\displaystyle{ 1.33*\sin(60) = 1.0*\sin(\theta_2) }[/math]

[math]\displaystyle{ \sin(\theta_2) = 1.33*\sin(60) = }[/math] 1.15

BUT...if we take arcsin of 1.15 we get an ERROR, because 1.15 is out of range of the sin function.

This means that there is no possible angle for this ray to make in the air. What if the ray is at an angle of 60 degrees to the vertical tries to emerge from the pool of water into the air above?

It doesn't emerge, but it reflects back down into the water! this is called total internal reflection.

Let's try to find the angle to the vertical for a ray in the water that gives a ray in the air at 90 degrees to the vertical. (Because 90 degrees is the largest possible angle from the vertical that is normal to the surface)

[math]\displaystyle{ 1.33*\sin(\theta_1) = 1.0*\sin(90) }[/math]

[math]\displaystyle{ \sin(\theta_1) = \frac{1.0}{1.33} = }[/math]0.75

[math]\displaystyle{ \theta_1 = }[/math] 49 degrees

Any underwater rays at more than 49 degrees to the vertical cannot escape into the air, but are internally reflected back into the water.

Connectedness

1. How is this topic connected to something that you are interested in?
2. How is it connected to your major?
3. Is there an interesting industrial application?

History

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

See also

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

Further reading

Books, Articles or other print media on this topic

External links

Internet resources on this topic

References

This section contains the the references you used while writing this page