Refraction of Light & Snell's Law: Videos & Practice Problems

Light refraction occurs when light transitions between materials, changing speed and direction. This bending is described by Snell's law, represented as $n_1\sin \left({\theta }_1\right)=n_2\sin \left({\theta }_2\right)$. When light enters a medium with a higher index of refraction, it bends toward the normal, while it bends away when entering a medium with a lower index. Understanding these principles is crucial for solving problems related to light behavior in different materials.

Snell's Law

Light refraction is a phenomenon that occurs when light travels from one medium to another, causing it to change speed and direction. This bending of light is crucial for understanding how lenses work and is governed by Snell's Law. Snell's Law states that the product of the index of refraction of the first medium and the sine of the angle of incidence is equal to the product of the index of refraction of the second medium and the sine of the angle of refraction. This relationship can be expressed mathematically as:

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

In this equation, \( n_1 \) and \( n_2 \) are the indices of refraction for the first and second materials, respectively, while \( \theta_1 \) is the angle of incidence (the angle at which light hits the boundary) and \( \theta_2 \) is the angle of refraction (the angle at which light exits into the new medium).

When light enters a medium with a higher index of refraction (like air to water), it bends towards the normal, resulting in an angle of refraction that is smaller than the angle of incidence. For example, if light enters water from air at a 30-degree angle, the angle of refraction can be calculated using Snell's Law. Given that the index of refraction for air is approximately 1 and for water is about 1.33, the calculation yields an angle of refraction of approximately 22.1 degrees.

Conversely, when light exits a medium with a higher index of refraction to one with a lower index (like glass to air), it bends away from the normal. In this case, if light exits glass (with an index of refraction of 1.46) into air at the same 30-degree angle, the angle of refraction is found to be approximately 46.9 degrees, which is greater than the angle of incidence.

Understanding these principles of light refraction and the application of Snell's Law is essential for solving problems related to optics, such as determining how light behaves as it passes through different materials. This knowledge is foundational for fields such as physics, engineering, and various applied sciences.

A laser pointer emits a ray which enters a quartz crystal at an angle 50° with the normal to the surface of the crystal. The ray bends inside the crystal, making an angle of 30° with the normal. Find the index of refraction of quartz.

• A. 1.00
• B. 0.65
• C. 1.53
• D. 3.77

Answer: (C)

Example 1

In this problem, we analyze the behavior of a light ray as it travels from air into a glass block, focusing on the concepts of refraction and the application of Snell's Law. The angle of incidence, denoted as θ₁, is given as 70 degrees. The indices of refraction for air (n₁) and glass (n₂) are 1.0 and 1.46, respectively. According to Snell's Law, the relationship between these angles and indices can be expressed as:

n₁ sin(θ₁) = n₂ sin(θ₂)

Substituting the known values, we have:

1.0 sin(70°) = 1.46 sin(θ₂)

Calculating sin(70°) gives approximately 0.94, leading to:

0.94 = 1.46 sin(θ₂)

From this, we can isolate sin(θ₂):

sin(θ₂) = 0.94 / 1.46

Taking the inverse sine, we find θ₂ to be approximately 40 degrees. This lower angle confirms the expected behavior of light bending towards the normal when entering a medium with a higher index of refraction.

Next, we determine θ_x, the angle with respect to the x-axis, which is complementary to θ₂:

θ_x = 90° - θ₂ = 90° - 40° = 50°.

To find the vertical distance (y) that the light travels within the glass before exiting, we utilize the tangent function, which relates the opposite side (y) to the adjacent side (3.75 cm, half the width of the glass block):

tan(θ_x) = y / 3.75

Substituting θ_x:

tan(50°) = y / 3.75

Rearranging gives:

y = 3.75 * tan(50°).

Calculating this yields y ≈ 4.47 cm. Thus, the light ray travels approximately 4.47 centimeters below the top of the glass block before exiting.

A ray of light is incident on a glass pane with an angle of 60°. The light partially reflects and partially refracts. What is the angle θ between the reflected and refracted rays?

• A. 66.4°
• B. 53.6°
• C. 36.4°
• D. 83.4°

Answer: (D)

Here’s what students ask on this topic:

What is Snell's Law and how is it used in refraction problems?

Snell's Law describes the relationship between the angles of incidence and refraction when light passes between two different media. It is given by the equation:

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Here, $n_1$ and $n_2$ are the indices of refraction for the two media, and ${\theta }_1$ and ${\theta }_2$ are the angles of incidence and refraction, respectively. This law is used to determine the angle at which light will refract when it enters a new medium, given the indices of refraction and the angle of incidence.

How does light behave when it enters a medium with a higher index of refraction?

When light enters a medium with a higher index of refraction, it slows down and bends toward the normal. This means the angle of refraction (${\theta }_2$) is smaller than the angle of incidence (${\theta }_1$). This behavior can be explained using Snell's Law:

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Since $n_2$ is greater than $n_1$, ${\theta }_2$ must be smaller to satisfy the equation.

What happens to light when it exits a medium with a higher index of refraction into a lower one?

When light exits a medium with a higher index of refraction into a lower one, it speeds up and bends away from the normal. This means the angle of refraction (${\theta }_2$) is larger than the angle of incidence (${\theta }_1$). Using Snell's Law:

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Since $n_1$ is greater than $n_2$, ${\theta }_2$ must be larger to satisfy the equation.

How do you calculate the angle of refraction using Snell's Law?

To calculate the angle of refraction using Snell's Law, you need the indices of refraction for both media and the angle of incidence. The law is given by:

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Rearrange the equation to solve for the angle of refraction (${\theta }_2$):

${\theta }_2=\sin -1\left(\frac{n_1\sin {\theta }_1}{n_2}\right)$

Plug in the known values for $n_1$, $n_2$, and ${\theta }_1$ to find ${\theta }_2$.

What is the index of refraction and how does it affect light refraction?

The index of refraction, denoted as $n$, is a measure of how much a material slows down light compared to a vacuum. It is defined as:

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

where $c$ is the speed of light in a vacuum and $v$ is the speed of light in the material. A higher index of refraction means light travels slower in that material, causing it to bend more when entering or exiting the material. This bending is described by Snell's Law.