Snell's Law
Snell's Law
Now that we know that the degree of bending, or the angle of refraction, is dependent on the refractive index of a medium, how do we calculate the angle of refraction? The answer to this question was discovered by a Dutch physicist called Willebrord Snell in 1621 and is now called Snell's Law or the Law of Refraction.
Definition: Snell's law
\[\boxed{n_1 \sin \theta_1 = n_2 \sin \theta_2}\] where \[\begin{array}{rl} n_1 &= \text{Refractive index of material 1} \\ n_2 &= \text{Refractive index of material 2} \\ \theta_1 &= \text{Angle of incidence} \\ \theta_2 &= \text{Angle of refraction} \\ \end{array}\]
Remember that angles of incidence and refraction are measured from the normal, which is an imaginary line perpendicular to the surface.
Suppose we have two media with refractive indices \(n_1\) and \(n_2\). A light ray is incident on the surface between these materials with an angle of incidence \(\theta_1\). The refracted ray that passes through the second medium will have an angle of refraction \(\theta_2\).
Fact:
Snell never published his discovery of the Law of Refraction. His work was actually published by another prominent physicist of the time, Christiaan Huygens, who gave credit to Snell.
A recommended project for formal assessment on verifying Snell's laws and determining the refractive index of an unknown transparent solid should be done. Learners will need a glass block, ray box, 0-360 protractor, a glass block, a transparent block of an unknown material and A4 paper. This project is given as two formal experiments, which learners can complete and then write up as a project. The project should include some background information on Snell's law, all the steps taken in the experiments, key experimental results and a conclusion about the experiment (including identification of the unknown solid from the refractive index).If \[n_2 > n_1\] then from Snell's Law, \[\sin \theta_1 > \sin \theta_2\]
For angles smaller than \(\text{90}\)\(\text{°}\), \(\sin{\theta}\) increases as \(\theta\) increases. Therefore, \[\theta_1 > \theta_2.\] This means that the angle of incidence is greater than the angle of refraction and the light ray is bent toward the normal.
Similarly, if \[n_2 < n_1\] then from Snell's Law, \[\sin \theta_1 < \sin \theta_2.\] For angles smaller than \(\text{90}\)\(\text{°}\), \(\sin{\theta}\) increases as \(\theta\) increases. Therefore, \[\theta_1 < \theta_2.\] This means that the angle of incidence is less than the angle of refraction and the light ray is bent away from the normal.
Light is moving from a medium with a higher refractive index to one with a lower refractive index. Light is refracted away from the normal.
Light is moving from a medium with a lower refractive index to a medium with a higher refractive index. Light is refracted towards the normal.
What happens to a ray that lies along the normal line? In this case, the angle of incidence is \(\text{0}\)\(\text{°}\) and \begin{align*} \sin \theta_2 &= \frac{n_1}{n_2} \sin \theta_1 \\ &= \text{0} \\ \therefore \theta_2 &= \text{0}. \end{align*}
This shows that if the light ray is incident at \(\text{0}\)\(\text{°}\), then the angle of refraction is also \(\text{0}\)\(\text{°}\).The direction of the light ray is unchanged, however, the speed of the light will change as it moves into the new medium. Therefore refraction still occurs although it will not be easily observed.
Let's use an everyday example that you can more easily imagine. Imagine you are pushing a lawnmower or a cart through short grass. As long as the grass is pretty much the same length everywhere, it is easy to keep the mower or cart going in a straight line. However, if the wheels on one side of the mower or cart enter an area of grass that is longer than the grass on the other side, that side of the mower will move more slowly since it is harder to push the wheels through the longer grass. The result is that the mower will start to turn inwards, towards the longer grass side as you can see in the picture. This is similar to light which passes through a medium and then enters a new medium with a higher refractive index or higher optical density. The light will change direction towards the normal, just like the mower in the longer grass. The opposite happens when the mower moves from an area of longer grass into an area with shorter grass. The side of the mower in the shorter grass will move faster andthe mower will turn outwards, just like a light ray moving from a medium with high refractive index into a medium with low refractive index moves away from the normal.
The lawnmower is moving from an area of short grass into an area of longer grass. When it reaches the boundary between the areas, the wheels which are in the long grass move slower than the wheels which are still in the short grass, causing the lawnmower to change direction and its path to bend inwards.
Example: Using Snell's Law
Question
Light is refracted at the boundary between water and an unknown medium. If the angle of incidence is \(\text{25}\)\(\text{°}\), and the angle of refraction is \(\text{20.6}\)\(\text{°}\), calculate the refractive index of the unknown medium and use the table from the previous lesson to identify the material.
Step 1: Determine what is given and what is being asked
The angle of incidence \(\theta_{1}=\)\(\text{25}\)\(\text{°}\)
The angle of refraction \(\theta_{2}=\)\(\text{20.6}\)\(\text{°}\)
We can look up the refractive index for water in the table from the previous lesson: \(n_{1} =\text{1.333}\)
We need to calculate the refractive index for the unknown medium and identify it.
Step 2: Determine how to approach the problem
We can use Snell's law to calculate the unknown refractive index, \(n_{2}\)
According to Snell's Law: \begin{align*} n_1 \sin \theta_1 &= n_2 \sin \theta_2 \\ n_2 &= \frac{n_1 \sin \theta_1}{\sin \theta_2} \\ n_2 &= \frac{\text{1.333} \sin \text{25}\text{°}}{\sin \text{20.6}\text{°}} \\ n_2 &= \text{1.6} \end{align*}
Step 3: Identify the unknown medium
According to the table from the previous lesson, typical glass has a refractive index between \(\text{1.5}\) to \(\text{1.9}\). Therefore the unknown medium is typical glass.
Example: Using Snell's Law
Question
A light ray with an angle of incidence of \(\text{35}\)\(\text{°}\) passes from water to air. Find the angle of refraction using Snell's Law and the table from the previous lesson. Discuss the meaning of your answer.
Step 1: Determine the refractive indices of water and air
From the table from the previous lesson, the refractive index is \(\text{1.333}\) for water and about \(\text{1}\) for air. We know the angle of incidence, so we are ready to use Snell's Law.
Step 2: Substitute values
According to Snell's Law: \begin{align*} n_1 \sin \theta_1 &= n_2 \sin \theta_2 \\ \text{1.33} \sin \text{35}\text{°} &= \text{1} \sin \theta_2 \\ \sin \theta_2 &= \text{0.763} \\ \theta_2 &= \text{49.7}\text{°} \text{ or } \text{130.3}\text{°} \end{align*}
Since \(\text{130.3}\)\(\text{°}\) is larger than \(\text{90}\)\(\text{°}\), the solution is: \[\theta_2 = \text{49.7}\text{°}\]
Step 3: Discuss the answer
The light ray passes from a medium of high refractive index to one of low refractive index. Therefore, the light ray is bent away from the normal.
Example: Using Snell's Law
Question
A light ray passes from water to diamond with an angle of incidence of \(\text{75}\)\(\text{°}\). Calculate the angle of refraction. Discuss the meaning of your answer.
Step 1: Determine the refractive indices of water and air
From the table from the previous lesson, the refractive index is \(\text{1.333}\) for water and \(\text{2.42}\) for diamond. We know the angle of incidence, so we are ready to use Snell's Law.
Step 2: Substitute values and solve
According to Snell's Law: \begin{align*} n_1 \sin \theta_1 &= n_2 \sin \theta_2 \\ \text{1.33} \sin \text{75}\text{°} &= \text{2.42} \sin \theta_2 \\ \sin{\theta_2} &= \text{0.531} \\ \theta_2 & = \text{32.1}\text{°} \end{align*}
Step 3: Discuss the answer
The light ray passes from a medium of low refractive index to one of high refractive index. Therefore, the light ray is bent towards the normal.
This lesson is part of:
Optics and Optical Phenomena