90-r 90- In the figure the second angle of incidence (Zis) = prism-material as 1.6 and Zi= 45°) 60⁰ 19⁰ 24⁰ 34⁰ (Take the refractive index of the
90-r 90- In the figure the second angle of incidence (Zis) = prism-material as 1.6 and Zi= 45°) 60⁰ 19⁰ 24⁰ 34⁰ (Take the refractive index of the
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![### Physics - Refraction Through a Prism
**Diagram Explanation:**
The provided diagram shows a triangular prism with an apex angle of 60°. A light ray is incident upon one face of the prism at an angle of \( 45^\circ \) (labeled as \( i \)). The light ray refracts at the point of incidence, passing through the prism, and exits out of the other face. The angles inside the prism indicate the relationship between the incident ray, the refracted ray, and the emergent ray.
**Problem Statement:**
In the figure, calculate the second angle of incidence (\(∠i_s\)). (Take the refractive index of the prism-material as 1.6 and \(∠i = 45^\circ\))
**Answer Choices:**
- \( 60^\circ \)
- \( 19^\circ \)
- \( 24^\circ \)
- \( 34^\circ \)
To solve this problem, use Snell's Law, which is expressed as:
\[ n_1 \sin i_1 = n_2 \sin r_1 \]
Given:
- Refractive index of the prism-material (\( n_2 \)): 1.6
- Refractive index of air (\( n_1 \)): 1
- Incident angle (\( i_1 \)): \( 45^\circ \)
First, calculate the angle of refraction (\( r_1 \)) inside the prism using Snell's Law:
\[ \sin r_1 = \frac{n_1 \sin i_1}{n_2} = \frac{1 \cdot \sin 45^\circ}{1.6} \approx \frac{0.7071}{1.6} \approx 0.442 \]
\[ r_1 \approx \sin^{-1}(0.442) \approx 26.2^\circ \]
The relationship between the angles inside the prism is given by:
\[ ϴ = A + r_1 + r_2 \]
For a symmetric configuration within a triangular prism:
\[ A = 60^\circ \]
And since:
\[ r_1 + r_2 = ϴ - A \]
By symmetry and internal consistency:
\[ r_1 = r_2 = r \approx 26.2^\circ \]
Then, angle i_s can be found directly](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8c939cff-bc46-4066-b22a-d4e55e2c89ab%2F20fb5697-0244-459c-a51a-90345edd0cd6%2Fzd1xzho_processed.png&w=3840&q=75)
Transcribed Image Text:### Physics - Refraction Through a Prism
**Diagram Explanation:**
The provided diagram shows a triangular prism with an apex angle of 60°. A light ray is incident upon one face of the prism at an angle of \( 45^\circ \) (labeled as \( i \)). The light ray refracts at the point of incidence, passing through the prism, and exits out of the other face. The angles inside the prism indicate the relationship between the incident ray, the refracted ray, and the emergent ray.
**Problem Statement:**
In the figure, calculate the second angle of incidence (\(∠i_s\)). (Take the refractive index of the prism-material as 1.6 and \(∠i = 45^\circ\))
**Answer Choices:**
- \( 60^\circ \)
- \( 19^\circ \)
- \( 24^\circ \)
- \( 34^\circ \)
To solve this problem, use Snell's Law, which is expressed as:
\[ n_1 \sin i_1 = n_2 \sin r_1 \]
Given:
- Refractive index of the prism-material (\( n_2 \)): 1.6
- Refractive index of air (\( n_1 \)): 1
- Incident angle (\( i_1 \)): \( 45^\circ \)
First, calculate the angle of refraction (\( r_1 \)) inside the prism using Snell's Law:
\[ \sin r_1 = \frac{n_1 \sin i_1}{n_2} = \frac{1 \cdot \sin 45^\circ}{1.6} \approx \frac{0.7071}{1.6} \approx 0.442 \]
\[ r_1 \approx \sin^{-1}(0.442) \approx 26.2^\circ \]
The relationship between the angles inside the prism is given by:
\[ ϴ = A + r_1 + r_2 \]
For a symmetric configuration within a triangular prism:
\[ A = 60^\circ \]
And since:
\[ r_1 + r_2 = ϴ - A \]
By symmetry and internal consistency:
\[ r_1 = r_2 = r \approx 26.2^\circ \]
Then, angle i_s can be found directly
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