To solve the problem of finding the angle of refraction of light as it passes from crown glass to water, we will use Snell's Law, which states:
\(\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]\)
where:
- \(\( n_1 \)\) is the refractive index of the first medium (glass in this case, \(\( n_1 = 1.52 \)\)).
- \(\( \theta_1 \)\) is the angle of incidence in the first medium (35°).
- \(\( n_2 \)\) is the refractive index of the second medium (water in this case, \(\( n_2 = 1.33 \)\)).
- \(\( \theta_2 \)\) is the angle of refraction in the second medium, which we need to find.
### Step-by-Step Solution:
1. Convert the Angle of Incidence to Radians:
\(\[
\theta_1 = 35^\circ
\]\)
To convert degrees to radians, use the conversion factor:
\(\[
\theta_1 = 35^\circ \times \frac{\pi}{180^\circ} = \frac{35\pi}{180} \approx 0.6109 \text{ radians}
\]\)
2. Apply Snell's Law:
\(\[
n_1 \sin(\theta_1) = n_2 \sin(\theta_2)
\]\)
Substitute the known values into Snell's Law:
\(\[
1.52 \sin(0.6109) = 1.33 \sin(\theta_2)
\]\)
3. Calculate \(\( \sin(\theta_1) \)\):
Using the sine function:
\(\[
\sin(0.6109) \approx 0.5736
\]\)
So,
\(\[
1.52 \times 0.5736 \approx 0.8719
\]\)
Thus, Snell's Law now looks like:
\(\[
0.8719 = 1.33 \sin(\theta_2)
\]\)
4. Solve for \(\( \sin(\theta_2) \)\):
\(\[
\sin(\theta_2) = \frac{0.8719}{1.33} \approx 0.6555
\]\)
5. Find \(\( \theta_2 \)\) Using the Inverse Sine Function:
\(\[
\theta_2 = \arcsin(0.6555)
\]\)
Using the arcsine function:
\(\[
\theta_2 \approx 40.91^\circ
\]\)
6. Round to the Nearest Tenth:
\(\[
\theta_2 \approx 40.9^\circ
\]\)
So, the angle of the corresponding refracted ray with respect to the normal is approximately \(\( 40.9^\circ \)\).
