Sure, let's tackle this problem step-by-step.
Given:
- Refractive index of water, [tex]\( n_1 = 1.33 \)[/tex]
- Refractive index of air, [tex]\( n_2 = 1.00 \)[/tex]
- Angle in water, [tex]\( \theta_1 = 35^\circ \)[/tex]
We need to determine the angle [tex]\( \theta_2 \)[/tex] when the light wave passes from water into air. This requires applying Snell's law, which is given by:
[tex]\[ \theta_2 = \sin^{-1} \left( \frac{n_1 \sin(\theta_1)}{n_2} \right) \][/tex]
Let's substitute the given values into the formula.
[tex]\[ \theta_2 = \sin^{-1} \left( \frac{1.33 \sin(35^\circ)}{1.00} \right) \][/tex]
To proceed, follow these steps:
1. Calculate [tex]\( \sin(35^\circ) \)[/tex]:
[tex]\[ \sin(35^\circ) \approx 0.5736 \][/tex]
2. Multiply [tex]\( n_1 \)[/tex] with [tex]\( \sin(35^\circ) \)[/tex]:
[tex]\[ n_1 \sin(35^\circ) = 1.33 \times 0.5736 \approx 0.7629 \][/tex]
3. Divide by [tex]\( n_2 \)[/tex]:
[tex]\[ \frac{0.7629}{1.00} = 0.7629 \][/tex]
4. Finally, calculate [tex]\( \sin^{-1}(0.7629) \)[/tex] to find [tex]\( \theta_2 \)[/tex]:
[tex]\[ \theta_2 = \sin^{-1}(0.7629) \approx 49.7167^\circ \][/tex]
Therefore, the angle [tex]\( \theta_2 \)[/tex] when the light passes from water to air is approximately [tex]\( 49.7167^\circ \)[/tex]. This matches the given answer in the options, rounded as [tex]\( 49.7^\circ \)[/tex].
Thus, the correct answer is:
B. [tex]\( 49.7^\circ \)[/tex]
