# Section1The Refraction of Light¶ permalink

In this series of experiments, we will begin to explore another feature of light found in geometric optics, namely, the ability of light rays to refract upon entering a transparent medium with a differing index of refraction. We will determine a mathematical relationship between the angle of incidence and the angle of refraction. Armed with the law of refraction, one can create lenses en route to technologies such as the microscope and telescope.

*We will always measure angles from the normal to the surface, unless explicitly noted otherwise.*

# Subsection1.1Refraction from Air to Water¶ permalink

Fill the semi-cylindrical, transparent plastic box nearly full of water and align it on a piece of polar graph paper resting on cardboard. Make sure the bottom of the vertical scratch (the center of curvature) on the straight side of the box falls on the intersection of the 0° - 180° and 90° - 270° lines on the paper. Stick a pin vertically on the 0° line (the “normal” to the surface) about 5 cm from the box.

Look at the pin through the water from the curved side of the box. Move your head until the pin and the vertical scratch on the box are in line. Mark this line of sight with another pin placed on your side of the box.

From this, we can conclude that when light passes from air to water at an angle of incidence of 0° there is no change in direction of the light. In other words, when the light enters normal to the box it experiences no "bending".

Change the position of the first pin to obtain an angle of incidence of 10°. With a second pin, mark the path through the water of the light from the first pin to the vertical scratch on the box. Measure the angle of refraction.

Repeat this procedure every 10° for angles of incidence up to about 80°.

\(\theta_{i}\) | \(\theta_{r}\) |

0 | 0 |

10 | 7 |

20 | 15 |

30 | 22 |

40 | 28 |

50 | 35 |

60 | 40 |

70 | 45 |

80 | N/A |

In each case, the resulting ray is refracted towards the normal, resulting in a smaller angle of refraction than the initial angle of incident. This is always true when rays are moving from air to water due to water having a higher index of refraction than air.

We can search, at various angles of incidence, for, in addition to the refracted ray, a ray reflected off the planar surface of the box. This reflection can be difficult, or impossible, to see for some angles.

At larger angles of incident, it is easier to see the ray reflected off of the box as reflection and refraction are inversely related. The incoming light is split between reflection and refraction, and thus as one gets stronger the other gets weaker. As such, when the incoming light travels along the normal there is no reflection as it it entirely refracted.

While \(\theta_r\) plots nicely as a function of \(\theta_i\text{,}\) we see that \(\sin\theta_r\) as a function of \(\sin\theta_i\) is more accurate. We can thus express, for this setup, the relationship as \(\theta_r = 0.75 \cdot \theta_i\text{.}\) As we know that the index of refraction for air is essentially \(1\text{,}\) we can determine that the index of refraction for water is \(1.\overline{3}\) (which we get from multiplying by \(0.75^{-1}\) on each side of our equation to get the standard form \(n_r\sin\theta_r = n_i\sin\theta_i\)).

We have assumed throughout this experiment that the results obtained by looking at the object through the water from the curved side are the same as if we were looking at the object from within the water; this assumption can be justified by noting that rays traveling along the normal exhibit no change due to refraction, and as we are dealing with rays passing though the center of the circle each outbound ray travels along a radius, and all radii are normal to the circular surface. This would not be the case if we were to use a rectangular of polygonal box as we would have to account for the double refraction.

# Subsection1.2Refraction from Water to Air¶ permalink

Align the semi-cylindrical box in the same manner as described in Section 1.1, but reverse the orientation of the curved side; ensure that the center of curvature is located at the origin of the polar coordinate system. Stick a pin vertically on the 0° line about 5 cm from the curved surface.

Look at the pin through the water from the straight side of the box. Move your head until the pin and the vertical scratch on the box are in line. Mark this line of sight with another pin. Measure the corresponding angles of refraction.

Repeat this procedure every 10° starting at 0° to about 60°.

\(\theta_{i}\) | \(\theta_{r}\) |

0 | 0 |

10 | 12 |

20 | 25 |

30 | 40 |

40 | 60 |

50 | 85 |

60 | N/A |

In each case, the resulting ray is refracted away from the normal, resulting in a larger angle of refraction than the initial angle of incident. This is always true when rays are moving from water to air due to water having a higher index of refraction than air.

When you reach a large enough angle of incidence, you can no longer see a refracted ray. For this experiment, this occurs a little after 50°. After this point, the incident light rays are totally reflected, resulting in the lack of refraction. Due to the inverse relation between reflection and refraction, at 0° the reflection is nonexistent as there is total refraction, and as we approach our critical point the reflection slowly strengthens while the refraction weakens. This can be accounted for by noting that our refractive interface is the planar face of our container which is a semi-reflective material, and as the reflection occurs inside the water it is more noticeable.

We can now demonstrate the reversibility of this relationship by performing this same experiment using our our refracted angles from Table Figure as our incidence angles. We can note from this that there is a relation in both directions that can be accounted for using our \(\sin\) relationship.

\(\theta_{i}\) | \(\theta_{r}\) |

0 | 0 |

7 | 10 |

15 | 20 |

22 | 30 |

28 | 40 |

35 | 50 |

40 | 60 |

45 | 70 |

From this, we can conclude that the relationship between the refraction from air to water and from water to air are reversible. This is where the idea of notating the refraction relationship between \(\theta_i\) and \(\theta_r\) as \(n_r\sin\theta_r = n_i\sin\theta_i\) where \(n_i\) and \(n_r\) are the indices of refraction corresponding to the mediums \(\theta_i\) and \(\theta_r\) are measured in.

So, we can determine the mathematical relationship between \(\theta_i\) and \(\theta_r\) during the refraction of light passing from water to air as \(\theta_r = (0.75^{-1}) \cdot \theta_i\text{.}\)

Using our mathematical relationship between \(\theta_i\) and \(\theta_r\text{,}\) we can determine the “critical angle,” the angle at which the light ray experiences no refraction as it is entirely reflected. To do so, we will solve for \(\theta_i\) when we set \(\theta_r\) to 90°.Doing this, we get that the maximum \(\theta_i\) that still exhibits some refraction is 48.59°, after this point the light ray is entirely reflected. This situation, total internal reflection, only occurs when light travels to a less dense medium as when light moves to a denser medium the light is bent towards the normal and thus even when the angle of incidence approaches 90° it will be in the visible region, even if it is highly distorted. This decrease in visibility is the reason that we were unable to experimentally determine the position of the 80° image as it was visible, but was so distorted that we were unable to determine the actual location.

Similar to when dealing with refraction from Air to Water, we can assume that the results are essentially the same as if we placed the object in the water due as we are measuring the rays from the center of curvature, and thus the initial ray until it hits the planar surface of the box is traveling along a radial path, and thus along the normal which means that it encounters no change in direction due to refraction.

# Subsection1.3Additional Activities¶ permalink

While exploring the properties of refraction between air and water we found a constant in both directions that allowed us to predict the refracted angle measurements. This constant is called the “index of refraction” and it is a property of the mediums involved. To show that this is a property of the medium and not a general constant we will find the index of refraction of a solid plastic D-shaped block.

We have determined our plastic block to have an index of refraction of 1.4, which we might expect due to it being more dense than water and thus will bend the light rays more than water did.

We can use our knowledge of refraction to consider the difference in actual position and apparent position of an object under water. By using Snell's law we determine that \(D_\text{apparent} = D_\text{real}\cdot\frac{n_a}{n_r}\) which for water gives \(D_\text{apparent} = \frac{4}{3}\cdot D_\text{real}\)

Using this relationship we determined the index of refraction of a heavy leaded glass block to be \(2.\overline{3}\text{.}\)