To solve the problem of determining the angle of refraction when a ray of light passes from ethanol to air given the specified incident angle and refractive indices, we can use Snell's Law. Let's go through it step-by-step.
Given:
- Incident angle, [tex]\(\theta_1 = 34^\circ\)[/tex]
- Refractive index of ethanol, [tex]\(n_1 = 1.36\)[/tex]
- Refractive index of air, [tex]\(n_2 = 1.00\)[/tex]
Snell's Law:
[tex]\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \][/tex]
Steps to find the angle of refraction ([tex]\(\theta_2\)[/tex]):
1. Convert the incident angle to radians:
[tex]\[ \theta_1 = 34^\circ = 34 \times \frac{\pi}{180} \text{ radians} \approx 0.5934 \text{ radians} \][/tex]
2. Apply Snell's Law:
[tex]\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \][/tex]
Therefore:
[tex]\[ \sin(\theta_2) = \frac{n_1 \sin(\theta_1)}{n_2} = \frac{1.36 \sin(0.5934)}{1.00} \][/tex]
3. Calculate [tex]\(\sin(\theta_1)\)[/tex]:
[tex]\[ \sin(0.5934) \approx 0.5592 \][/tex]
4. Find [tex]\(\sin(\theta_2)\)[/tex]:
[tex]\[ \sin(\theta_2) = 1.36 \times 0.5592 \approx 0.7605 \][/tex]
5. Find [tex]\( \theta_2 \)[/tex] by taking the inverse sine (arcsin):
[tex]\[ \theta_2 = \sin^{-1}(0.7605) \][/tex]
Convert this result back to degrees:
[tex]\[ \theta_2 \approx 49.5085^\circ \][/tex]
Choosing the closest answer from the given options:
- A. [tex]\( 21^\circ \)[/tex]
- B. [tex]\( 34^\circ \)[/tex]
- C. [tex]\( 50^\circ \)[/tex]
- D. [tex]\( 66^\circ \)[/tex]
The calculated angle of refraction is approximately [tex]\(49.5085^\circ\)[/tex], which is closest to option C ([tex]\(50^\circ\)[/tex]).
Therefore, the correct answer is C. [tex]\(50^\circ\)[/tex].
