A beam of light shines on an equilateral glass prism at an...

A beam of light shines on an equilateral glass prism at an angle of $45^{\circ}$ to one face (Eigure 23-47a). (a) What is the angle at which the light emerges from the opposite face given that nglass $=1.57^{n_{\text {glass }}}=1.57$ (b) Now consider what happens when dispersion is involved ( Figure 23-47b). Assume the incident ray of light spans the spectrum of visible light between 400 and 700 $\mathrm{nm}$ (violet to red, respectively). The index of refraction for violet light in the glass prism is 1.572, and it is $1.568$ for red light in the glass prism. Find the distance along the right face of the prism between the points where the red light and violet light emerge back into air. Assume the prism is $10.0 \mathrm{~cm}$ on a side and the incident ray hits the midpoint of the left face. Example $23-3$

Key Concepts

Prism Geometry

Prism geometry involves understanding how the shape and angles of a prism affect the path of light as it is refracted at its surfaces. The internal angles of the prism, along with the orientation at which light enters and exits, determine the overall angular deviation of the light ray, leading to effects such as minimum deviation and even total internal reflection under certain conditions.

Dispersion

Dispersion is the phenomenon in which the refractive index of a material varies with the wavelength of light, causing different colors to bend by different amounts when passing through a glass medium. This wavelength-dependent behavior is essential in explaining how a prism can separate white light into its constituent colors, as shorter and longer wavelengths refract differently.

Refraction

Refraction is the bending of light as it passes from one medium into another where its speed is different. This change in direction is governed by the variation in the optical density of the two media and is described by Snell’s law, which relates the angles of incidence and refraction to the refractive indices of the media.

Snell’s Law

Snell’s law is a fundamental principle in optics that quantifies the relationship between the angles of incidence and refraction when light crosses between different media. It states that the product of the refractive index and the sine of the light’s angle in one medium is equal to the corresponding product in another medium, providing a mathematical framework to calculate the direction of light in various materials.

Chromatic Dispersion Effects

Chromatic dispersion effects refer to the differences in the refractive behavior of light at various wavelengths. As the index of refraction changes with color, light composed of multiple wavelengths will emerge from a prism at slightly different angles. This separation is key to understanding phenomena such as the formation of rainbows and the precise prediction of emerging light positions in dispersive optical applications.

Transcript

00:01 For this problem on the topic of wave properties of light, we're told that a beam of light is shining on an equilateral glass prism at an angle of 45 degrees to one face, as shown in figure a.

00:11 We want to know the angle at which the light emerges from the opposite face, given that the refractive index of glass is 1 .57.

00:18 We then want to consider what happens when dispersion is involved, and we want to find the distance along the right face of the prism between the points where the red light and violet light emerge back into a.

00:30 Now we've drawn the ray diagram for part a as above, and the diffraction at the a glass interface can be represented as by snell's law, sine 1, sine theta 1 equal to n2, sine theta 2, which means that sine theta 2 is equal to n1 over n2 times sine theta 1, and theta 2 is equal to the arc sign of n1 over n2 times sine theta 1.

01:19 Putting in our values, we get this to be the arc sign of 1 over 1 .57 times the sign of 45 degrees, which is an angle theta 2 of 26 .77 degrees.

01:40 And now from geometry we get beta to be 90 degrees minus 26 .77 degrees, which is 63 .23 degrees, gamma equal to 100 and 80 degrees minus 60 degrees, which gives us 56 .77.

02:14 Degrees and theta 3 equal to 90 degrees minus 56 .77 degrees which gives us theta 3 to be 33 .23 degrees.

02:31 Now refraction at the glass air interface gives us n3 sine theta 3 equal to n4 sine theta 4.

02:44 And so rearranging we get theta 4 to be the arc sign of n3 over n4 times the sign of theta 3, which is the arc sign of 1 .57 divided by 1 times the sign of 33 .23 degrees.

03:14 This gives us an angle theta 4 of 59 .4 degrees.

03:22 Now for part 4, we'll, for part b rather, we'll draw the ray diagram again, which looks as follows.

03:33 And for a red light, we have the refractive index of a times the sign of theta 1 is equal to the defective index for the red light times the sign of theta 1 is equal to the defective index for the red light times the sign of the angle theta for red light.

03:51 And so we get the angle of refraction for red light theta red equal to the arc sign of the refractive index of a divided by the refractive index for the red light in red times the sign of theta 1...

Please give Ace some feedback