Problem 44 - University Physics with Modern Physics

Question

A light ray of wavelength 700 . \(\mathrm{nm}\) traveling in air \(\left(n_{1}=1.00\right)\) is incident on a boundary with a liquid \(\left(n_{2}=1.63\right) .\) a) What is the frequency of the refracted ray? b) What is the speed of the refracted ray? c) What is the wavelength of the refracted ray?

Step-by-Step Solution

Answer

Question: A light ray with an initial wavelength of 700 nm travels from air (refractive index 1.00) into a liquid (refractive index 1.63). Calculate the frequency, speed, and wavelength of the refracted ray in the liquid. Answer: The refracted ray has a frequency of \(4.286\times10^{14}\, \mathrm{Hz}\), a speed of \(1.840\times10^8\, \mathrm{m/s}\) in the liquid, and a wavelength of \(430\, \mathrm{nm}\) in the liquid.

Step 1: Calculate the initial speed and frequency of the light ray in air

Recalling that the refractive index is given by \(n = \frac{c}{v}\), where \(v\) is the speed of light in the medium. We can solve for the speed of light in air as: $$ v_{1} = \frac{c}{n_{1}} $$ Then, we can calculate the frequency of the light ray using the relationship \(v = f \lambda\), where \(f\) is the frequency, and \(\lambda\) is the wavelength: $$ f = \frac{v_{1}}{\lambda_{1}} $$

Step 2: Calculate the speed of the refracted ray in the liquid

Using the refractive index \(n_{2}\), we find the speed of the refracted light ray in the liquid as: $$ v_{2} = \frac{c}{n_{2}} $$

Step 3: Calculate the wavelength of the refracted ray in the liquid

Since the frequency is constant when light travels from one medium to another, we have: $$ f_{1} = f_{2} $$ And by using the relationship \(v = f \lambda\), we can calculate the wavelength of the refracted ray in the liquid as: $$ \lambda_{2} = \frac{v_{2}}{f_{2}} = \frac{v_{2}}{f_{1}} $$ Now let's plug in the given values and solve for the required quantities:

Step 4: Plug in the given values and solve

From steps 1, 2, and 3, we have: $$ v_{1} = \frac{3.00\times10^8\, \mathrm{m/s}}{1.00} = 3.00\times10^8\, \mathrm{m/s} $$ $$ f = \frac{3.000\times10^8\, \mathrm{m/s}}{700\times10^{-9}\, \mathrm{m}} = 4.286\times10^{14}\, \mathrm{Hz} $$ $$ v_{2} = \frac{3.00\times10^8\, \mathrm{m/s}}{1.63} = 1.840\times10^8\, \mathrm{m/s} $$ $$ \lambda_{2} = \frac{1.840\times10^8\, \mathrm{m/s}}{4.286\times10^{14}\, \mathrm{Hz}} = 430\times10^{-9}\, \mathrm{m} $$ So, the refracted ray has: a) a frequency of \(4.286\times10^{14}\, \mathrm{Hz}\), b) a speed of \(1.840\times10^8\, \mathrm{m/s}\) in the liquid, c) a wavelength of \(430\, \mathrm{nm}\) in the liquid.

Key Concepts

Refractive Index

The refractive index is a fundamental concept in optics that describes how light propagates through different media. When light travels from one medium to another, such as from air to water, it changes speed and direction due to the difference in the refractive index of the two media.
The refractive index (\(n\)) is defined as the ratio of the speed of light in a vacuum (\(c\)) to the speed of light in the medium (\(v\)). Mathematically, it can be expressed as:

• \(n = \frac{c}{v}\)

A higher refractive index indicates that light travels slower in the medium compared to air (\(n=1.00\) for air). For example, in the provided exercise, the liquid has a refractive index (\(n_2\)) of 1.63, indicating light slows down considerably compared to its speed in a vacuum.
The relationship is key because a higher refractive index implies a greater bending of the light ray upon entering the medium. This bending, or refraction, is what causes the wavelength and speed of the light to change, while the frequency remains constant.

Wavelength Change

When light enters a medium with a different refractive index, its speed changes, leading to a change in wavelength. However, the frequency of light does not change. In simple terms, the number of wave crests passing a point per second remains the same when light moves between different media.
The relationship between speed, frequency, and wavelength in any medium is given by:

• \(v = f \lambda\)

Here, \(v\) stands for the speed of light, \(f\) is the frequency, and \(\lambda\) is the wavelength. When light moves into a medium where it slows down (\(n > 1\)), the wavelength decreases. In our exercise, the wavelength in air is 700 nm, while it reduces to around 430 nm in the liquid. This decrease happens because even though the speed of light decreases, the frequency remains constant, necessitating a proportional decrease in wavelength. Understanding this concept helps explain why objects appear differently when viewed under water or through transparent substances with a different refractive index.

Light Speed in a Medium

The speed of light is not constant across different media. In a vacuum, light travels at its maximum speed of\(3.00 \times 10^8\, \mathrm{m/s}\). However, this speed decreases when light passes through other materials. How much it slows down depends on the material's refractive index.
The rule of thumb is: the higher the refractive index, the slower the light travels in that medium. This is depicted as:

• \(v = \frac{c}{n}\)

In the example provided, light travels in air with a refractive index of 1.00, maintaining its full speed. When it enters the liquid with a refractive index of 1.63, the speed of light is reduced to \(1.840 \times 10^8\, \mathrm{m/s}\). This slowing down results in the change of other properties like wavelength, as seen before. By understanding how light speed varies in different materials, we gain insights into phenomena such as refraction, which is crucial in lenses and other optical instruments, and even in natural phenomena like the bending of light through a prism to form a rainbow.