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$\bullet$$\bullet$ Use the graph in Figure 23.29 for silicate flint glass. (a) What are the indexes of refraction of this glass for extreme violet light of wavelength 400 $\mathrm{nm}$ and for extreme red light of wavelength 700 $\mathrm{nm}$ ? (b) What are the wavelengths of 400 nm violet light and 700 $\mathrm{nm}$ red light in this glass? (c) Calculate the ratio of the speed of extreme red light to that of extreme violet light in the glass. Which of these travels faster in the glass? (d) If a beam of white light in air strikes a sheet of this glass at $65.0^{\circ}$ with the normal in air, what will be the angle of dispersion between the extremes of visible light in the glass? In other words, what will be the angle between extreme red and extreme violet light in the glass?

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a) for violet light, the index of refraction is 1.66 and for red light, the index of refraction is 1.61b) Wavelength of violet light is 241 $\mathrm{nm}$ light is $[435 \mathrm{nm}]$c) $1.03$red$1.16^{\circ}$

Physics 102 Electricity and Magnetism

Physics 103

Chapter 23

Electromagnetic Waves and Propagationof Light

Electromagnetic Waves

Reflection and Refraction of Light

Rutgers, The State University of New Jersey

Simon Fraser University

Hope College

Lectures

02:30

In optics, ray optics is a geometric optics method that uses ray tracing to model the propagation of light through an optical system. As in all geometric optics methods, the ray optics model assumes that light travels in straight lines and that the index of refraction of the optical material remains constant throughout the system.

10:00

In optics, reflection is the change in direction of a wavefront at an interface between two different media so that the wavefront returns into the medium from which it originated. Common examples include the reflection of light, sound and water waves. The law of reflection says that for specular reflection the angle at which the wave is incident on the surface equals the angle at which it is reflected. Reflection may also be referred to as "mirror image" or "specular reflection". Refraction is the change in direction of a wave due to a change in its speed. The refractive index of a material is a measure of its ability to change the direction of a wave. A material with a higher refractive index will change the direction of a wave to a greater degree than a material with a lower refractive index. When a wave crosses the boundary between two materials with different refractive indices, part of the wave is refracted; that is, it changes direction. The ratio of the speeds of propagation of the two waves determines the angle of refraction, which is the angle between the direction of the incident and the refractive rays.

02:15

Use the graph in Figure 23…

02:03

The index of refraction fo…

09:15

A thin beam of white light…

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A narrow beam of white lig…

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The table gives the index …

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A hydrogen discharge lamp …

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$\bullet$ A narrow beam of…

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A light beam containing re…

01:56

In the first part of this problem, we have to find the index of refraction for 2 for the given wave lengths so from the graph we can see that the index of reflection for red light is an r and it is equal to 1.61. Similarly, for what light it is and- and it is equal to 1.66 now in part b- we have to calculate the wavelength of red and voilet light in glass. So for a light we can write lambda r equals to lambda divided by r. Here this wanda is the wavelength of red light in glass landais the wave length of the light in vacuum. The reflective index now inserting the values into this equation. We can write, it is lambda r equals to we have the value for lambda is 700 nanometer divided by we have the n r as 1.61. So this will give us a value for lambda r is lambda r equals to 435 manometer. Similarly, the vivent for volelight in glass will be lamda v equals to lambda, divided by n v, where this lambda is the wavelength of this wadelinvacuum. So by inserting the values into the square, we can write 400 nanometer divided by. We have the value for n v as 1.66, so lamda v will be equals to 241 nanometer. Now, in part c of this problem, we have to calculate the ratio of speed of red light to the ratio speed of olelight. So we call this ratio, as we are divided by v v, where this v r is the speed of red light and v v is the speed of boiling light, as we know that the retractantx is written as an equal to c divided by v. Where c is the speed of light and a hum, and is the speed of light in that material? So for a light, we can write it as n r equals to c divided by v r. This coin can be written as we are equal to c divided by n r. We call this equation as equation number 1. Similarly, we can write in entherequation is v v that is equal to c divided by n v. We call this equation as equation: number 2. On now dividing equation: number 1 by equation number 2: we can write, we are divided by v v. This should be equals to n divided by n r. We call this equation as equation: number 3. Now inserting the values into the quai we can write v r divided by v v equals to 1.66 divided by 1.61, so this is equal to 1.03. So this is the ratio, so this ratio shows that this v r is greater than v v. Now, in part of this problem, we have to calculate the angle of dispersion between 2 extremes of visible light, so this angle is theta. In order to calculate this angle, let me apply the senate's law as a n r, sine theta r equals to n a sine theta a so we can write. This equation for theta is theta equal to r side of n a divided by r, and there is sine theta a. We call this equation as equation: number 5. Now inserting the values into this coin, we can write thet r equals to arc sine. We have the value for n, is 1.00 divided by n r is equal to 1.61 and that sine theta is equal to 65.0 degree. So this will give us a value for theta is theta equals to 34.26 degrees. Now theta can be written as a theta equals to outside of n a divided by n v, and then there will be sine theta v. The sine theta is equals to sine theta a and we call the square in his equation. Number 6. Now, by inserting values into the square, we can write it as theta v equals to ride of 1.00 divided by 1.66, and then there will be sine of 65.0 degree. So this will give us the value for theta v, as theta v equals to 33.09 degree. It now we can write the value for theta is theta equals to theta r minus theta v. So by writing these the values into this question. Theta equals to 34.26 minus 26 degree minus 33.09 degree. So this will give us the value for this. Theta is theta equals to 1.16 degree, so this is the required answer. Thank you.

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