Sure, let's go through this step by step.
Step 1: Identify the known values.
- The index of refraction for glass, [tex]\( n_{\text{glass}} = 1.50 \)[/tex]
- The index of refraction for water, [tex]\( n_{\text{water}} = 1.33 \)[/tex]
- The angle of incidence in the glass, [tex]\( \theta_{\text{glass}} = 35^\circ \)[/tex]
Step 2: Use Snell's Law to find the angle of refraction in water.
Snell's Law is given by:
[tex]\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \][/tex]
Rearranging to solve for [tex]\(\theta_2\)[/tex]:
[tex]\[ \theta_2 = \sin^{-1}\left(\frac{n_1 \sin(\theta_1)}{n_2}\right) \][/tex]
Plug in the known values:
[tex]\[ \theta_{\text{water}} = \sin^{-1}\left(\frac{1.50 \cdot \sin(35^\circ)}{1.33}\right) \][/tex]
Step 3: Calculation.
Evaluate the expression inside the sine inverse function:
- First, compute [tex]\(\sin(35^\circ)\)[/tex]. Let's convert 35 degrees to radians and find [tex]\(\sin(35^\circ)\)[/tex].
- [tex]\(\sin(35^\circ) \approx 0.5736\)[/tex]
- Now, calculate the ratio [tex]\(\frac{1.50 \cdot 0.5736}{1.33} \approx 0.6468\)[/tex]
Next, use the inverse sine (arcsine) function to find the angle whose sine is approximately 0.6468.
[tex]\[
\theta_{\text{water}} = \sin^{-1}(0.6468) \approx 40.3^\circ
\][/tex]
Step 4: Conclusion.
Based on the calculations, the angle of refraction as the light wave passes from glass into water is approximately [tex]\(40.3^\circ\)[/tex].
Therefore, the correct answer is:
C. [tex]\(40.3^\circ\)[/tex]
