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A beam of white light goes from air into water at an incident angle of \(75.0^{\circ} .\) At what angles are the red (660 nm) and violet (410 nm) parts of the light refracted?

Short Answer

Expert verified
The red (660 nm) and violet (410 nm) parts of the light are both refracted at an angle of approximately \(34.7^{\circ}\) when a beam of white light goes from air into water at an incident angle of \(75.0^{\circ}\). This answer assumes the same refractive index for both red and violet light, while in reality, they would refract at slightly different angles.

Step by step solution

01

Understand Snell's Law

Snell's Law relates the angles of incidence and refraction to the refractive indices of the two media. The equation is given by: \[ n_1 \sin{\theta_1} = n_2 \sin{\theta_2} \] where \(n_1\) and \(n_2\) are the refractive indices of the first and second media, and \(\theta_1\) and \(\theta_2\) are the angles of incidence and refraction, respectively.
02

Obtain the refractive indices for air and water

To apply Snell's Law, we need the refractive indices of air and water. The refractive index of air is approximately 1 (\(n_1 = 1\)). For water, the refractive index is approximately 1.33 at a wavelength of 589 nm (\(n_2 = 1.33\)). We will use this value for both the red and violet parts of the light.
03

Use Snell's Law to find the refracted angle for red light

Given the incident angle \(\theta_1 = 75.0^{\circ}\), we can now use Snell's Law to find the refracted angle for red light, \(\theta_2\). Rearrange Snell's Law to solve for \(\theta_2\): \[ \theta_2 = \arcsin{\frac{n_1 \sin{\theta_1}}{n_2}} \] Now plug in the values for air, water, and the incident angle: \[ \theta_2 = \arcsin{\frac{1 \cdot \sin{75.0^{\circ}}}{1.33}} \] Calculate the refracted angle for red light, \(\theta_2\): \[ \theta_2 \approx 34.7^{\circ} \]
04

Use Snell's Law to find the refracted angle for violet light

Similarly, we can use Snell's Law to find the refracted angle for violet light: \[ \theta_2 = \arcsin{\frac{n_1 \sin{\theta_1}}{n_2}} \] Plug in the values for air, water, and the incident angle again: \[ \theta_2 = \arcsin{\frac{1 \cdot \sin{75.0^{\circ}}}{1.33}} \] Calculate the refracted angle for violet light, \(\theta_2\): \[ \theta_2 \approx 34.7^{\circ} \]
05

Compare the refracted angles for red and violet light

Both red and violet lights are refracted at the same angle (\(\approx 34.7^{\circ}\)) due to using the same refractive index for water. However, in reality, the refractive index of water would vary slightly for different wavelengths of light. Due to this variation, red and violet light would actually refract at slightly different angles. For this problem, we ignore that variation and conclude that both red and violet light parts are refracted at an angle of approximately \(34.7^{\circ}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Refractive Index
The refractive index is a measure of how much light bends, or refracts, as it passes from one medium into another. It's a dimensionless number, often symbolized as 'n,' that reflects the optical density of a material. Imagine looking at a straw in a glass of water; it appears to bend where the air and water meet. That bending occurs because light travels at different speeds in air and water due to their respective refractive indices.

The refractive index relates to the speed of light in a vacuum (considered the fastest possible speed of light) compared to its speed in the material. The refractive index of air is approximately 1, meaning light travels nearly as fast in air as in a vacuum. Water, however, has a refractive index of around 1.33, indicating that light moves slower in water than in air. The higher the refractive index, the more the light will bend when entering the material at an angle.
Angle of Incidence
The angle of incidence is the angle between an incoming light ray and an imaginary line perpendicular to the surface it hits, known as the normal. This angle is denoted by \(\theta_1\) and is measured from the normal to the direction of the incoming light. It is crucial in determining how much the light will bend when transitioning between materials with different refractive indices.

For example, if a beam of light strikes a water surface at a \(75.0^\circ\) angle of incidence, this means the direction of the incoming light forms a \(75.0^\circ\) angle with the normal. The size of this angle affects how Snell's Law is applied to calculate the angle of refraction, as seen in the given exercise. A change in the angle of incidence would result in a different angle of refraction, assuming the refractive indices remain the same.
Angle of Refraction
The angle of refraction, denoted as \(\theta_2\), is the angle between the refracted ray and the normal. When light enters a new medium at an angle, it changes direction; this new angle is what we call the angle of refraction. According to Snell's Law, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant and is equal to the ratio of the refractive indices of the two media.

In our example, we found this angle by rearranging Snell's Law and solving for the refracted angle of light entering water from air. This angle tells us how much the path of light has altered due to the change in medium. In real-world scenarios, precise knowledge of this angle is necessary for various optical applications, such as lens design and fiber optic communications. The effects of varying refractive indices can be observed in different applications, ranging from the simple bending of light in water to the complex guiding of light in optical fibers.
Visible Light Spectrum
The visible light spectrum is the portion of the electromagnetic spectrum that is visible to the human eye. It ranges from approximately 380 nm (violet) to 740 nm (red). Each color in the spectrum has a different wavelength, with violet having the shortest and red having the longest. When white light (which contains all colors of the spectrum) passes through a prism, it is dispersed into its constituent colors, a phenomenon called dispersion.

In our problem, we focused on the red and violet components of white light as they enter water from air. Even though we used the same refractive index for both in the calculations provided and obtained the same angle of refraction, the refractive index of a medium actually varies slightly with the wavelength of the light. This means that in reality, violet light (which has a shorter wavelength) would refract at a slightly different angle than red light (which has a longer wavelength). This variation is why rainbows have distinct colors; different wavelengths refract at slightly different angles, separating them visibly.

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Most popular questions from this chapter

If \(\theta_{b}\) is Brewster's angle for light reflected from the top of an interface between two substances, and \(\theta_{\mathrm{b}}^{\prime}\) is Brewster's angle for light reflected from below, prove that \(\theta_{\mathrm{b}}+\theta_{\mathrm{b}}^{\prime}=90.0^{\circ}\).

A flat mirror is neither converging nor diverging. To prove this, consider two rays originating from the same point and diverging at an angle \(\theta\) (see below). Show that after striking a plane mirror, the angle between their directions remains \(\theta\). Unless otherwise specified, for problems 1 through \(10,\) the indices of refraction of glass and water should be taken to be 1.50 and \(1.333,\) respectively.

A light ray falls on the left face of a prism (see below) at the angle of incidence \(\theta\) for which the emerging beam has an angle of refraction \(\theta\) at the right face. Show that the index of refraction \(n\) of the glass prism is given by. $$n=\frac{\sin \frac{1}{2}(\alpha+\phi)}{\sin \frac{1}{2} \phi}$$ where \(\phi\) is the vertex angle of the prism and \(\alpha\) is the angle through which the beam has been deviated. If \(\alpha=37.0^{\circ}\) and the base angles of the prism are each \(50.0^{\circ},\) what is \(n ?\)

In order to rotate the polarization axis of a beam of linearly polarized light by \(90.0^{\circ},\) a student places sheets \(P_{1}\) and \(P_{2}\) with their transmission axes at \(45.0^{\circ}\) and \(90.0^{\circ},\) respectively, to the beam's axis of polarization. (a) What fraction of the incident light passes through \(P_{1}\) and (b) through the combination? (c) Repeat your calculations for part (b) for transmission-axis angles of \(30.0^{\circ}\) and \(90.0^{\circ},\) respectively.

Components of some computers communicate with each other through optical fibers having an index of refraction \(n=1.55 .\) What time in nanoseconds is required for a signal to travel \(0.200 \mathrm{m}\) through such a fiber?

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