at night, a light under the surface of a pool shines up the surface of the water (n = 1.33) at an angle of 32.0°.
what is the angle of refraction once it hits the air?

40.4°
44.8°
47.0°
38.2°

Question

at night, a light under the surface of a pool shines up the surface of the water (n = 1.33) at an angle of 32.0°.
what is the angle of refraction once it hits the air?

40.4°
44.8°
47.0°
38.2°

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": true, "concepts_used": [ "Snells Law", "Refractive Index" ], "new_concepts": [], "current_concepts": [ "Snells Law", "Refractive Index", "Light Refraction" ] } ## Identify the given parameters Using the *Refractive Index* knowledge point $$ \begin{aligned} &n_1 = 1.33 \quad \text{(water)}\ &n_2 = 1.00 \quad \text{(air)}\ &\theta_1 = 32.0^\circ \quad \text{(angle of incidence)} \end{aligned} $$ ## Apply Snell's Law to find the angle of refraction Using the *Snells Law* knowledge point $$ \begin{aligned} &n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\ &1.33 \cdot \sin(32.0^\circ) = 1.00 \cdot \sin(\theta_2)\ &\sin(\theta_2) = 1.33 \cdot 0.5299 = 0.7048\ &\theta_2 = \arcsin(0.7048) \approx 44.8^\circ \end{aligned} $$

(A) $40.4^\circ$

(B) $44.8^\circ$

(C) $47.0^\circ$

(D) $38.2^\circ$

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QUESTION IMAGE

Question

Explanation:

Identify the given parameters

Apply Snell's Law to find the angle of refraction

Answer: