To determine the angle of the light wave when it passes from water (with refractive index [tex]\( n_1 = 1.33 \)[/tex]) into air (with refractive index [tex]\( n_2 = 1.00 \)[/tex]), we can use Snell's Law, which is given by the equation:
[tex]\[
\theta_2 = \sin^{-1}\left(\frac{n_1 \sin (\theta_1)}{n_2}\right)
\][/tex]
Here, [tex]\(\theta_1\)[/tex] is the angle of incidence in water, which is given as [tex]\(35^\circ\)[/tex].
Step-by-step solution:
1. Determine the known values:
- Refractive index of water, [tex]\( n_1 = 1.33 \)[/tex]
- Refractive index of air, [tex]\( n_2 = 1.00 \)[/tex]
- Incident angle, [tex]\( \theta_1 = 35^\circ \)[/tex]
2. Convert the angle [tex]\(\theta_1\)[/tex] from degrees to radians,
since trigonometric functions in the formula require the angle to be in radians.
[tex]\[
\theta_1 = 35^\circ = \theta_1 \times \frac{\pi}{180} = 35 \times \frac{\pi}{180} \approx 0.611 \text{ radians}
\][/tex]
3. Calculate the sine of the incident angle [tex]\(\theta_1\)[/tex]:
[tex]\[
\sin(\theta_1) = \sin(0.611) \approx 0.573
\][/tex]
4. Apply Snell’s Law to find [tex]\(\theta_2\)[/tex]:
[tex]\[
\theta_2 = \sin^{-1}\left(\frac{n_1 \sin (\theta_1)}{n_2}\right) = \sin^{-1}\left(\frac{1.33 \times 0.573}{1.00}\right) = \sin^{-1}(0.7629)
\][/tex]
5. Calculate the value inside the inverse sine function and then the angle [tex]\(\theta_2\)[/tex]:
[tex]\[
\theta_2 = \sin^{-1}(0.7629) \approx 0.868 \text{ radians}
\][/tex]
6. Convert [tex]\(\theta_2\)[/tex] from radians back to degrees:
[tex]\[
\theta_2 = 0.868 \text{ radians} \times \frac{180}{\pi} \approx 49.7^\circ
\][/tex]
Therefore, the angle of the light wave when it passes from water into air is approximately [tex]\( 49.7^\circ \)[/tex].
The correct answer is:
C. [tex]\( 49.7^\circ \)[/tex]
