Question

A ray of light traveling in water is incident on an interface with a flat piece of glass. The wavelength of the light in the water is 726 nm, and its wavelength in the glass is 544 nm. If the ray in water makes an angle of 56.0\(^\circ\) with respect to the normal to the interface, what angle does the refracted ray in the glass make with respect to the normal?

Short answer

The refracted angle in the glass is approximately 39.2°.

Step-by-step solution

Understand Snell's Law

The relationship between the angle of incidence and the angle of refraction when light passes from one medium to another is given by Snell's Law: \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \), where \( n_1 \) and \( n_2 \) are the refractive indices of the first and second mediums, respectively, and \( \theta_1 \) and \( \theta_2 \) are the angles of incidence and refraction.

Calculate the Refractive Index Using Wavelength

The refractive index of a medium for a certain wavelength can be calculated by comparing it with the speed of light in that medium. It is also inversely proportional to the wavelength in that medium. Thus, the ratio of the wavelengths in water and glass gives the ratio of the refractive indices: \( \frac{\lambda_1}{\lambda_2} = \frac{n_2}{n_1} \). Here, \( \lambda_1 = 726 \) nm for water and \( \lambda_2 = 544 \) nm for glass.

Solve for the Ratio of Refractive Indices

Using the given wavelengths, the ratio of the refractive indices can be calculated: \( n_{water}/n_{glass} = 544/726 \approx 0.749 \).

Apply Snell's Law to Find the Angle in Glass

Using Snell's Law and the derived refractive index ratio, we can substitute into the equation: \( 1 \cdot \sin(56.0^\circ) = 0.749 \cdot \sin(\theta_2) \). Calculate \( \sin(\theta_2) \).

Calculate the Angle of Refraction

Solve for \( \theta_2 \) by taking the inverse sine of \( \sin(\theta_2) \). Compute \( \theta_2 = \arcsin\left( \frac{\sin(56.0^\circ)}{0.749} \right) \approx 39.2^\circ \).

Key concepts

Refractive Index

The refractive index, often symbolized as "n," is a crucial concept when studying how light behaves as it moves between different materials. The refractive index quantifies how much light slows down in a medium compared to its speed in a vacuum.

• A high refractive index means light travels more slowly through the medium.
• The value is determined by comparing the speed of light in the medium to the speed of light in a vacuum, or via wavelength changes since the speed and wavelength are inversely related: the shorter the wavelength, the higher the refractive index.

In the given problem, we used the refractive index to determine how light bends as it travels from water to glass, and vice versa.

Angle of Incidence

The angle of incidence refers to the angle between an incoming ray of light and a line perpendicular to the surface at the point of contact, called the normal.

• This is the angle from where the light originates, such as the water in the exercise example.
• Angle of incidence is one of the critical factors Snell's Law uses to calculate the angle of refraction when light transfers to a different medium.

In the exercise, the angle of incidence was given as 56.0° in the water before transitioning to glass, setting the stage for determining how much light bends.

Angle of Refraction

The angle of refraction is the angle between the refracted ray and the normal line in the new medium. It tells us how much the path of light bends as it crosses into a new material.

• This angle is calculated using Snell's Law, which connects the refractive indices of the two media with the angles of incidence and refraction.
• For the light moving from water to glass in our problem, knowing the angle of refraction helps us understand the light's new direction.

Through the steps of applying trigonometric functions, we can solve for the angle of refraction, which was found to be approximately 39.2° for this example.

Wavelength Change

When light enters a different medium, its speed and wavelength change, although its frequency remains constant. This change in wavelength is essential as it impacts how much light bends.

• The wavelength in a medium is inversely proportional to its refractive index; as the refractive index increases, the wavelength decreases.
• In the exercise, light's wavelength changed from 726 nm in water to 544 nm in glass, meaning the light slowed down, causing the path to bend due to differing refractive indices.

Understanding changes in wavelength helps in solving how light will behave when it transitions from one medium to another.

Medium Transition

Medium transition occurs when light moves from one material, or medium, into another, such as from water into glass.

• This transition can lead to changes in speed, direction, and wavelength, depending on each medium's refractive index.
• Snell’s Law gives a reliable means to predict these changes by comparing the refractive indices of the two media involved.

For the scenario in the problem, transitioning from water to glass caused the light to bend in a measurable way due to differences in both refractive index and wavelength. Understanding medium transition is crucial for optics and explains phenomena like refraction.