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To solve for the angle of refraction when a ray of light passes from air into glass, we'll use Snell's Law. Snell's Law relates the angle of incidence and angle of refraction to the refractive indices of the two media.
The formula for Snell's Law is:
[tex]\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \][/tex]
Where:
- [tex]\( n_1 \)[/tex] is the refractive index of the first medium (air in this case, with [tex]\( n_1 = 1.00 \)[/tex]).
- [tex]\( \theta_1 \)[/tex] is the angle of incidence (30°).
- [tex]\( n_2 \)[/tex] is the refractive index of the second medium (glass in this case, with [tex]\( n_2 = 1.50 \)[/tex]).
- [tex]\( \theta_2 \)[/tex] is the angle of refraction, which we need to find.
Here's the step-by-step process:
1. Identify the values given:
- Angle of incidence, [tex]\( \theta_1 \)[/tex]: [tex]\(30^\circ\)[/tex]
- Refractive index of air, [tex]\( n_1 \)[/tex]: 1.00
- Refractive index of glass, [tex]\( n_2 \)[/tex]: 1.50
2. Convert the angle of incidence from degrees to radians for calculation:
[tex]\[ \theta_1 = 30^\circ \][/tex]
[tex]\[ \theta_1 \text{ in radians} = 0.5236 \text{ rad} \][/tex]
3. Apply Snell's Law:
[tex]\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \][/tex]
[tex]\[ 1.00 \cdot \sin(0.5236) = 1.50 \cdot \sin(\theta_2) \][/tex]
4. Solving for [tex]\( \sin(\theta_2) \)[/tex]:
[tex]\[ \sin(\theta_2) = \frac{1.00 \cdot \sin(0.5236)}{1.50} \][/tex]
[tex]\[ \sin(\theta_2) = \frac{\sin(0.5236)}{1.50} \][/tex]
[tex]\[ \sin(\theta_2) = \frac{0.5}{1.50} \][/tex]
[tex]\[ \sin(\theta_2) \approx 0.3333 \][/tex]
5. Find [tex]\( \theta_2 \)[/tex] by taking the arcsine (inverse sine) of [tex]\(0.3333\)[/tex]:
[tex]\[ \theta_2 \approx \arcsin(0.3333) \][/tex]
[tex]\[ \theta_2 \approx 0.3398 \text{ rad} \][/tex]
6. Convert [tex]\( \theta_2 \)[/tex] from radians back to degrees:
[tex]\[ \theta_2 \approx 19.4712^\circ \][/tex]
Therefore, the angle of refraction of the light as it enters the glass is approximately [tex]\(19.47^\circ\)[/tex].
The formula for Snell's Law is:
[tex]\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \][/tex]
Where:
- [tex]\( n_1 \)[/tex] is the refractive index of the first medium (air in this case, with [tex]\( n_1 = 1.00 \)[/tex]).
- [tex]\( \theta_1 \)[/tex] is the angle of incidence (30°).
- [tex]\( n_2 \)[/tex] is the refractive index of the second medium (glass in this case, with [tex]\( n_2 = 1.50 \)[/tex]).
- [tex]\( \theta_2 \)[/tex] is the angle of refraction, which we need to find.
Here's the step-by-step process:
1. Identify the values given:
- Angle of incidence, [tex]\( \theta_1 \)[/tex]: [tex]\(30^\circ\)[/tex]
- Refractive index of air, [tex]\( n_1 \)[/tex]: 1.00
- Refractive index of glass, [tex]\( n_2 \)[/tex]: 1.50
2. Convert the angle of incidence from degrees to radians for calculation:
[tex]\[ \theta_1 = 30^\circ \][/tex]
[tex]\[ \theta_1 \text{ in radians} = 0.5236 \text{ rad} \][/tex]
3. Apply Snell's Law:
[tex]\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \][/tex]
[tex]\[ 1.00 \cdot \sin(0.5236) = 1.50 \cdot \sin(\theta_2) \][/tex]
4. Solving for [tex]\( \sin(\theta_2) \)[/tex]:
[tex]\[ \sin(\theta_2) = \frac{1.00 \cdot \sin(0.5236)}{1.50} \][/tex]
[tex]\[ \sin(\theta_2) = \frac{\sin(0.5236)}{1.50} \][/tex]
[tex]\[ \sin(\theta_2) = \frac{0.5}{1.50} \][/tex]
[tex]\[ \sin(\theta_2) \approx 0.3333 \][/tex]
5. Find [tex]\( \theta_2 \)[/tex] by taking the arcsine (inverse sine) of [tex]\(0.3333\)[/tex]:
[tex]\[ \theta_2 \approx \arcsin(0.3333) \][/tex]
[tex]\[ \theta_2 \approx 0.3398 \text{ rad} \][/tex]
6. Convert [tex]\( \theta_2 \)[/tex] from radians back to degrees:
[tex]\[ \theta_2 \approx 19.4712^\circ \][/tex]
Therefore, the angle of refraction of the light as it enters the glass is approximately [tex]\(19.47^\circ\)[/tex].
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