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:
[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 (glass in this case, [tex]\( n_1 = 1.52 \)[/tex]).
- [tex]\( \theta_1 \)[/tex] is the angle of incidence in the first medium (35°).
- [tex]\( n_2 \)[/tex] is the refractive index of the second medium (water in this case, [tex]\( n_2 = 1.33 \)[/tex]).
- [tex]\( \theta_2 \)[/tex] 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:
[tex]\[
\theta_1 = 35^\circ
\][/tex]
To convert degrees to radians, use the conversion factor:
[tex]\[
\theta_1 = 35^\circ \times \frac{\pi}{180^\circ} = \frac{35\pi}{180} \approx 0.6109 \text{ radians}
\][/tex]
2. Apply Snell's Law:
[tex]\[
n_1 \sin(\theta_1) = n_2 \sin(\theta_2)
\][/tex]
Substitute the known values into Snell's Law:
[tex]\[
1.52 \sin(0.6109) = 1.33 \sin(\theta_2)
\][/tex]
3. Calculate [tex]\( \sin(\theta_1) \)[/tex]:
Using the sine function:
[tex]\[
\sin(0.6109) \approx 0.5736
\][/tex]
So,
[tex]\[
1.52 \times 0.5736 \approx 0.8719
\][/tex]
Thus, Snell's Law now looks like:
[tex]\[
0.8719 = 1.33 \sin(\theta_2)
\][/tex]
4. Solve for [tex]\( \sin(\theta_2) \)[/tex]:
[tex]\[
\sin(\theta_2) = \frac{0.8719}{1.33} \approx 0.6555
\][/tex]
5. Find [tex]\( \theta_2 \)[/tex] Using the Inverse Sine Function:
[tex]\[
\theta_2 = \arcsin(0.6555)
\][/tex]
Using the arcsine function:
[tex]\[
\theta_2 \approx 40.91^\circ
\][/tex]
6. Round to the Nearest Tenth:
[tex]\[
\theta_2 \approx 40.9^\circ
\][/tex]
So, the angle of the corresponding refracted ray with respect to the normal is approximately [tex]\( 40.9^\circ \)[/tex].
