A ray of monochromatic light of frequency 5.09x10 hertz is traveling from water into medium X. The angle of incidence in water is 45 and the angle of refraction in medium X is 29°, as show 12 (a) Normal 45° Water Medium X 29° Calculate the absolute index of refraction of X. Show all work, including the equation and substitution with units
A ray of monochromatic light of frequency 5.09x10 hertz is traveling from water into medium X. The angle of incidence in water is 45 and the angle of refraction in medium X is 29°, as show 12 (a) Normal 45° Water Medium X 29° Calculate the absolute index of refraction of X. Show all work, including the equation and substitution with units
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Refraction of Monochromatic Light**
When a ray of monochromatic light with a frequency of 5.09 x 10^14 Hz travels from a medium of water into another medium (labeled as X), the light undergoes refraction at the interface of the two media.
In this scenario:
- The angle of incidence in water is 45°.
- The angle of refraction in medium X is 29°, as depicted in the diagram below.
**Diagram of Refraction**:
- A ray of light is shown traveling from the water into medium X.
- The interface between the water and medium X is indicated by a horizontal line.
- The incident ray in the water medium forms an angle of 45° with the normal (a dashed vertical line).
- The refracted ray in medium X forms an angle of 29° with the normal.
**Objective**:
Calculate the absolute index of refraction of medium X. Be sure to show all work, including the equations and substitution with units.
**Formula and Calculation**:
To find the index of refraction of medium X, we use Snell's Law, which is defined as:
\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]
In this equation:
- \( n_1 \) is the index of refraction of the first medium (water).
- \( \theta_1 \) is the angle of incidence in the first medium (45°).
- \( n_2 \) is the index of refraction of the second medium (medium X).
- \( \theta_2 \) is the angle of refraction in the second medium (29°).
Given:
- The index of refraction of water, \( n_1 \), is approximately 1.33.
We rearrange the formula to solve for \( n_2 \) (index of refraction of medium X):
\[ n_2 = \frac{n_1 \sin(\theta_1)}{\sin(\theta_2)} \]
Substitute the known values into the equation:
\[ n_2 = \frac{1.33 \sin(45°)}{\sin(29°)} \]
Using the sine values:
\[ \sin(45°) \approx 0.707 \]
\[ \sin(29°) \approx 0.484 \]
Now, substitute these values:
\[ n_](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F666a7050-d460-4d4f-9624-d4bdba52f50b%2F04f2d5f1-03a5-4ec4-a090-6de1f9489b47%2Fnnt2yej_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Refraction of Monochromatic Light**
When a ray of monochromatic light with a frequency of 5.09 x 10^14 Hz travels from a medium of water into another medium (labeled as X), the light undergoes refraction at the interface of the two media.
In this scenario:
- The angle of incidence in water is 45°.
- The angle of refraction in medium X is 29°, as depicted in the diagram below.
**Diagram of Refraction**:
- A ray of light is shown traveling from the water into medium X.
- The interface between the water and medium X is indicated by a horizontal line.
- The incident ray in the water medium forms an angle of 45° with the normal (a dashed vertical line).
- The refracted ray in medium X forms an angle of 29° with the normal.
**Objective**:
Calculate the absolute index of refraction of medium X. Be sure to show all work, including the equations and substitution with units.
**Formula and Calculation**:
To find the index of refraction of medium X, we use Snell's Law, which is defined as:
\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]
In this equation:
- \( n_1 \) is the index of refraction of the first medium (water).
- \( \theta_1 \) is the angle of incidence in the first medium (45°).
- \( n_2 \) is the index of refraction of the second medium (medium X).
- \( \theta_2 \) is the angle of refraction in the second medium (29°).
Given:
- The index of refraction of water, \( n_1 \), is approximately 1.33.
We rearrange the formula to solve for \( n_2 \) (index of refraction of medium X):
\[ n_2 = \frac{n_1 \sin(\theta_1)}{\sin(\theta_2)} \]
Substitute the known values into the equation:
\[ n_2 = \frac{1.33 \sin(45°)}{\sin(29°)} \]
Using the sine values:
\[ \sin(45°) \approx 0.707 \]
\[ \sin(29°) \approx 0.484 \]
Now, substitute these values:
\[ n_

Transcribed Image Text:**Question:**
(b) Based on your calculation from 3B.39, what material could Medium X possibly be made of?
**Answer:**
*Provide your answer in the space below.*
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