QuestionJuly 25, 2025

A light ray strikes the surface of water at an angle of 40^circ What is the angle at which it penetrates the water.

A light ray strikes the surface of water at an angle of 40^circ What is the angle at which it penetrates the water.
A light ray strikes the surface of water at an angle of 40^circ What is the angle at which it penetrates the water.

Solution
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Answer

28.9^\circ Explanation 1. Identify Known Values Incident angle \theta_1 = 40^\circ, refractive index of air n_1 \approx 1.0, refractive index of water n_2 \approx 1.33. 2. Apply Snell's Law Use **Snell's Law**: n_1 \sin(\theta_1) = n_2 \sin(\theta_2) to find the refraction angle \theta_2. 3. Solve for Refraction Angle Rearrange to \sin(\theta_2) = \frac{n_1}{n_2} \sin(\theta_1). Calculate \sin(\theta_2) = \frac{1.0}{1.33} \sin(40^\circ). 4. Calculate and Convert to Degrees \sin(\theta_2) \approx 0.4695. Thus, \theta_2 \approx \arcsin(0.4695) \approx 28.9^\circ.

Explanation

1. Identify Known Values<br /> Incident angle $\theta_1 = 40^\circ$, refractive index of air $n_1 \approx 1.0$, refractive index of water $n_2 \approx 1.33$.<br />2. Apply Snell's Law<br /> Use **Snell's Law**: $n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$ to find the refraction angle $\theta_2$.<br />3. Solve for Refraction Angle<br /> Rearrange to $\sin(\theta_2) = \frac{n_1}{n_2} \sin(\theta_1)$. Calculate $\sin(\theta_2) = \frac{1.0}{1.33} \sin(40^\circ)$.<br />4. Calculate and Convert to Degrees<br /> $\sin(\theta_2) \approx 0.4695$. Thus, $\theta_2 \approx \arcsin(0.4695) \approx 28.9^\circ$.
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