13.3 Refraction

Refraction is the bending of light as it passes between different media. This phenomenon occurs due to changes in light speed, which varies depending on the medium's optical density. Understanding refraction is crucial for explaining everyday optical effects like why a straw appears bent in a glass of water or how lenses focus light.

The index of refraction quantifies how much light slows in a medium compared to its speed in a vacuum. Snell's law describes the relationship between angles of incidence and refraction when light crosses media boundaries. These concepts help explain total internal reflection and its applications in technologies like fiber optic communications.

Refraction of light between media

Change in light direction

When light travels from one transparent medium to another, it often changes direction at the boundary. This bending of light rays is what we call refraction.

• Refraction occurs because light travels at different speeds in different materials
• The amount of bending depends on both the difference in optical densities and the angle at which light strikes the boundary
• If light enters a denser medium (like going from air to water), it bends toward the normal line (perpendicular to the surface)
• If light enters a less dense medium (like going from water to air), it bends away from the normal line

Speed of light in media

Light travels at its maximum speed in a vacuum, but slows down when passing through matter. This change in speed is what causes refraction to occur.

• In vacuum, light travels at approximately 3 × 10⁸ m/s (commonly denoted as c)
• In air, light travels only slightly slower than in vacuum (about 99.97% of c)
• In water, light travels at about 75% of its vacuum speed
• In glass, light travels at about 67% of its vacuum speed
• In diamond, light slows to about 41% of its vacuum speed

The more a material slows down light, the more dramatic the bending effect when light enters or exits that material.

Index of refraction

The index of refraction provides a way to quantify how much a material slows down light compared to its speed in a vacuum.

$n=\frac{c}{v}$

Where:

• n is the index of refraction (dimensionless)
• c is the speed of light in vacuum (3 × 10⁸ m/s)
• v is the speed of light in the medium

Common indices of refraction include:

• Vacuum: exactly 1.0
• Air: approximately 1.0003
• Water: about 1.33
• Glass: typically 1.5-1.6
• Diamond: about 2.42

The higher the index of refraction, the more light slows down in that medium, and the more it will bend when entering or exiting that medium at an angle.

Snell's law

Snell's law provides the mathematical relationship between the angles of incidence and refraction when light passes between media with different indices of refraction.

$n_1 \sin \theta_1 = n_2 \sin \theta_2$

Where:

• n₁ is the index of refraction of the first medium
• θ₁ is the angle of incidence (measured from the normal)
• n₂ is the index of refraction of the second medium
• θ₂ is the angle of refraction (measured from the normal)

This law has several important implications:

• When light travels from a lower index to a higher index medium (like air to water), the refracted ray bends toward the normal (θ₂ < θ₁)
• When light travels from a higher index to a lower index medium (like water to air), the refracted ray bends away from the normal (θ₂ > θ₁)
• If light strikes the boundary along the normal (θ₁ = 0°), it continues straight through without bending (θ₂ = 0°)
• The greater the difference between the indices of refraction, the more pronounced the bending effect

Total internal reflection

Total internal reflection is a special case that occurs when light attempts to move from a higher index medium to a lower index medium at a sufficiently large angle of incidence.

• Total internal reflection only occurs when light travels from a higher index medium to a lower index medium (like water to air)
• It happens when the angle of incidence exceeds a specific value called the critical angle
• At the critical angle, the refracted ray would travel exactly along the boundary between the two media (at 90° to the normal)
• The critical angle can be calculated using:

$\theta_{\text{critical}} = \sin^{-1}\left(\frac{n_2}{n_1}\right)$

Where n₁ is the index of the medium the light is in, and n₂ is the index of the medium the light would enter.

When the angle of incidence exceeds the critical angle:

• No light passes into the second medium
• All light is reflected back into the first medium
• The reflection follows the law of reflection (angle of incidence equals angle of reflection)

Total internal reflection is the principle behind:

• Fiber optic communications, where light signals bounce along glass or plastic fibers
• The brilliant sparkle of diamonds, which have a high index of refraction
• Certain types of prisms used in binoculars and periscopes

Practice Problem 1: Index of Refraction

Solution

To find the index of refraction, we need to use the formula that relates the speed of light in vacuum to the speed of light in the medium:

$n = \frac{c}{v}$

Where:

• n is the index of refraction
• c is the speed of light in vacuum (3.00 × 10⁸ m/s)
• v is the speed of light in the medium (1.94 × 10⁸ m/s)

Substituting the values:

$n = \frac{3.00 \times 10^8 \text{ m/s}}{1.94 \times 10^8 \text{ m/s}} = 1.55$

Therefore, the index of refraction of the material is 1.55, which is typical of certain types of glass.

Practice Problem 2: Snell's Law

Solution

To find the angle of refraction, we'll use Snell's law:

$n_1 \sin \theta_1 = n_2 \sin \theta_2$

Where:

• n₁ = 1.00 (index of refraction of air)
• θ₁ = 42° (angle of incidence)
• n₂ = 1.33 (index of refraction of water)
• θ₂ = ? (angle of refraction, what we're solving for)

Rearranging to solve for θ₂:

$\sin \theta_2 = \frac{n_1 \sin \theta_1}{n_2} = \frac{1.00 \times \sin 42°}{1.33}$

$\sin \theta_2 = \frac{0.669}{1.33} = 0.503$

$\theta_2 = \sin^{-1}(0.503) = 30.2°$

Therefore, the light ray will refract at an angle of 30.2° to the normal in the water. Note that the angle decreased as the light entered the medium with the higher index of refraction, which is consistent with our understanding of refraction.

Practice Problem 3: Total Internal Reflection

Solution

To find the critical angle for total internal reflection, we use the formula:

$\theta_{\text{critical}} = \sin^{-1}\left(\frac{n_2}{n_1}\right)$

Where:

• n₁ = 2.42 (index of refraction of diamond)
• n₂ = 1.00 (index of refraction of air)

Substituting these values:

$\theta_{\text{critical}} = \sin^{-1}\left(\frac{1.00}{2.42}\right) = \sin^{-1}(0.413) = 24.4°$

Therefore, the critical angle for total internal reflection at a diamond-air interface is 24.4°. Any light ray inside the diamond that strikes the diamond-air boundary at an angle greater than 24.4° to the normal will undergo total internal reflection. This relatively small critical angle is why diamonds appear so brilliant - much of the light that enters a diamond is trapped inside by repeated total internal reflections before eventually exiting.