nair= 1.00 0₁ 0₁ Upper plastic-air boundary 021 nplastic In the picture above, a light ray travels from air and strikes a rectangular piece of plastic at an angle 0₁ = 42.0°. The ray then refracts into the plastic at an angle 0₁ = 29.0°. a) 8 point Solve for the index of refraction of the plastic. Show your work. b) 4 points Solve for the angle 02. Show your work. c) 12 Will this ray undergo total internal reflection in the plastic at the upper plastic-air boundary? Explain and show your work.
nair= 1.00 0₁ 0₁ Upper plastic-air boundary 021 nplastic In the picture above, a light ray travels from air and strikes a rectangular piece of plastic at an angle 0₁ = 42.0°. The ray then refracts into the plastic at an angle 0₁ = 29.0°. a) 8 point Solve for the index of refraction of the plastic. Show your work. b) 4 points Solve for the angle 02. Show your work. c) 12 Will this ray undergo total internal reflection in the plastic at the upper plastic-air boundary? Explain and show your work.
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A mirror is made of glass that is coated with a metal amalgam on one side due to which the light ray incident on the surface undergoes reflection and not refraction.
Question
![**Educational Content: Refraction and Total Internal Reflection**
**Diagram Explanation:**
The diagram shows a light ray traveling from air, with a refractive index (\( n_{\text{air}} = 1.00 \)), into a plastic material. The light initially strikes the plastic at an angle \(\theta_i = 42.0^\circ\) to the normal. Upon entering the plastic, it refracts, making an angle \(\theta_1 = 29.0^\circ\) with the normal. The refracted ray then continues within the plastic, eventually reaching the upper plastic-air boundary and potentially refracting again at angle \(\theta_2\).
**Problem Description:**
In the given setup, a light ray travels from the air into a rectangular plastic medium at an incident angle of \(\theta_i = 42.0^\circ\). Inside the plastic, it refracts at \(\theta_1 = 29.0^\circ\). Your tasks are:
**a) Solve for the index of refraction of the plastic. Show your work. (8 points)**
To solve this, you will apply Snell's Law:
\[ n_{\text{air}} \cdot \sin(\theta_i) = n_{\text{plastic}} \cdot \sin(\theta_1) \]
**b) Solve for the angle \(\theta_2\). Show your work. (4 points)**
Determine the angle of refraction \(\theta_2\) as the light exits back into the air using Snell's Law.
**c) Will this ray undergo total internal reflection in the plastic at the upper plastic-air boundary? Explain and show your work. (12 points)**
Analyze whether total internal reflection occurs, which happens if the angle of incidence inside the plastic is greater than the critical angle. Calculate the critical angle using:
\[ \sin(\theta_{\text{critical}}) = \frac{n_{\text{air}}}{n_{\text{plastic}}} \]
Explore these concepts thoroughly to understand the applications of Snell's Law and the conditions for total internal reflection.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F49f9d394-65f9-403f-909c-32dbefeb4065%2F2c37b26f-806a-4e55-8b69-bea219238f7e%2Fv0hsbj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Educational Content: Refraction and Total Internal Reflection**
**Diagram Explanation:**
The diagram shows a light ray traveling from air, with a refractive index (\( n_{\text{air}} = 1.00 \)), into a plastic material. The light initially strikes the plastic at an angle \(\theta_i = 42.0^\circ\) to the normal. Upon entering the plastic, it refracts, making an angle \(\theta_1 = 29.0^\circ\) with the normal. The refracted ray then continues within the plastic, eventually reaching the upper plastic-air boundary and potentially refracting again at angle \(\theta_2\).
**Problem Description:**
In the given setup, a light ray travels from the air into a rectangular plastic medium at an incident angle of \(\theta_i = 42.0^\circ\). Inside the plastic, it refracts at \(\theta_1 = 29.0^\circ\). Your tasks are:
**a) Solve for the index of refraction of the plastic. Show your work. (8 points)**
To solve this, you will apply Snell's Law:
\[ n_{\text{air}} \cdot \sin(\theta_i) = n_{\text{plastic}} \cdot \sin(\theta_1) \]
**b) Solve for the angle \(\theta_2\). Show your work. (4 points)**
Determine the angle of refraction \(\theta_2\) as the light exits back into the air using Snell's Law.
**c) Will this ray undergo total internal reflection in the plastic at the upper plastic-air boundary? Explain and show your work. (12 points)**
Analyze whether total internal reflection occurs, which happens if the angle of incidence inside the plastic is greater than the critical angle. Calculate the critical angle using:
\[ \sin(\theta_{\text{critical}}) = \frac{n_{\text{air}}}{n_{\text{plastic}}} \]
Explore these concepts thoroughly to understand the applications of Snell's Law and the conditions for total internal reflection.
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