The final angle of refraction in this case refractive index = 1.6). 34⁰ 64° 86⁰ 32⁰ (Take Zi= 45° and for the prism-material
The final angle of refraction in this case refractive index = 1.6). 34⁰ 64° 86⁰ 32⁰ (Take Zi= 45° and for the prism-material
Related questions
Question
![### Understanding Refraction through a Prism
#### Criteria:
- **Incident Angle (i)** = 45°
- **Refractive Index of Prism Material (n)** = 1.6
#### Question:
The final angle of **refraction** in this case = __________.
#### Options:
- 34°
- 64°
- 86°
- 32°
#### Explanation of Diagram:
The provided diagram illustrates the following key elements:
1. **Prism Shape**: An equilateral triangle is depicted.
2. **Angles**: The angle of deviation (60°), angle of incidence (i), and normal lines at the points where the light ray enters and exits the prism are marked. The refracted and emergent rays inside the prism demonstrate the bending due to the change in medium.
3. **Rays**:
- The incident ray enters the prism at an angle of 45°.
- The refracted ray bends inside the prism.
- The outgoing ray exits at a new angle (the final angle of refraction) which needs to be calculated.
#### Calculation Steps:
1. **Snell's Law Application at First Surface**:
\[ n_{\text{air}} \sin(i) = n_{\text{prism}} \sin(r) \]
Where \( n_{\text{air}} = 1 \), \( i = 45° \), \( n_{\text{prism}} = 1.6 \)
\[ \sin(45°) = 1.6 \sin(r) \]
\[ \frac{\sqrt{2}}{2} = 1.6 \sin(r) \]
\[ 0.707 = 1.6 \sin(r) \]
\[ \sin(r) = \frac{0.707}{1.6} \]
\[ r = \sin^{-1}(0.4418) \]
\[ r ≈ 26.43° \]
2. **Calculation at Second Surface**:
The internal angle at the second surface is:
\[ \text{Angle of prism} (60°) - r \]
\[ 60° - 26.43° = 33.57° \]
3. **Application of Snell’s Law again**:
\[ n_{\text{prism}} \sin(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8c939cff-bc46-4066-b22a-d4e55e2c89ab%2F171842c5-ab3f-4d8a-9c27-15f1f6e5eb31%2Ftktm7gr_processed.png&w=3840&q=75)
Transcribed Image Text:### Understanding Refraction through a Prism
#### Criteria:
- **Incident Angle (i)** = 45°
- **Refractive Index of Prism Material (n)** = 1.6
#### Question:
The final angle of **refraction** in this case = __________.
#### Options:
- 34°
- 64°
- 86°
- 32°
#### Explanation of Diagram:
The provided diagram illustrates the following key elements:
1. **Prism Shape**: An equilateral triangle is depicted.
2. **Angles**: The angle of deviation (60°), angle of incidence (i), and normal lines at the points where the light ray enters and exits the prism are marked. The refracted and emergent rays inside the prism demonstrate the bending due to the change in medium.
3. **Rays**:
- The incident ray enters the prism at an angle of 45°.
- The refracted ray bends inside the prism.
- The outgoing ray exits at a new angle (the final angle of refraction) which needs to be calculated.
#### Calculation Steps:
1. **Snell's Law Application at First Surface**:
\[ n_{\text{air}} \sin(i) = n_{\text{prism}} \sin(r) \]
Where \( n_{\text{air}} = 1 \), \( i = 45° \), \( n_{\text{prism}} = 1.6 \)
\[ \sin(45°) = 1.6 \sin(r) \]
\[ \frac{\sqrt{2}}{2} = 1.6 \sin(r) \]
\[ 0.707 = 1.6 \sin(r) \]
\[ \sin(r) = \frac{0.707}{1.6} \]
\[ r = \sin^{-1}(0.4418) \]
\[ r ≈ 26.43° \]
2. **Calculation at Second Surface**:
The internal angle at the second surface is:
\[ \text{Angle of prism} (60°) - r \]
\[ 60° - 26.43° = 33.57° \]
3. **Application of Snell’s Law again**:
\[ n_{\text{prism}} \sin(
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 3 images
