A man shines a flashlight from a boat into the water, illuminating a rock as in the figure below. What is the angle of incidence ₁ (in degrees)? HINT O 2.72 mi 02 -2.05 m-

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**Title: Calculating the Angle of Incidence - Snell’s Law Application**

**Description:**

In the presented scenario, a man is positioned on a boat, directing a flashlight into the water. The light illuminates a rock beneath the water's surface. The task is to calculate the angle of incidence, denoted as \(\theta_1\).

**Diagram Explanation:**

- The man on the boat shines a beam of light, represented as a blue arrow, onto a rock located underwater.
- The interface between the air and water is depicted as a horizontal line at the water's surface.
- The angle \(\theta_1\) is formed between the incident light ray and the normal (vertical dashed line) to the water surface.
- The angle \(\theta_2\) is formed between the refracted light ray (as it travels within the water) and the normal.
- Measurements provided:
  - The vertical distance from the water surface to the point of incidence is 2.72 meters.
  - The horizontal distance from the point of incidence to the rock is 2.05 meters.

**Hint Box:**
- A hint box is included with an input box for entering an angle in degrees.

To determine the angle of incidence \(\theta_1\), apply the concepts of refraction, taking into consideration Snell's Law and the geometry of the triangle formed by the light path.

\[ \text{Snell's Law: } n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]

Where:
- \(n_1\) is the refractive index of air (approx. 1.0)
- \(n_2\) is the refractive index of water (approx. 1.33)

Using these principles, solve for the angles using the provided diagram measurements.
Transcribed Image Text:**Title: Calculating the Angle of Incidence - Snell’s Law Application** **Description:** In the presented scenario, a man is positioned on a boat, directing a flashlight into the water. The light illuminates a rock beneath the water's surface. The task is to calculate the angle of incidence, denoted as \(\theta_1\). **Diagram Explanation:** - The man on the boat shines a beam of light, represented as a blue arrow, onto a rock located underwater. - The interface between the air and water is depicted as a horizontal line at the water's surface. - The angle \(\theta_1\) is formed between the incident light ray and the normal (vertical dashed line) to the water surface. - The angle \(\theta_2\) is formed between the refracted light ray (as it travels within the water) and the normal. - Measurements provided: - The vertical distance from the water surface to the point of incidence is 2.72 meters. - The horizontal distance from the point of incidence to the rock is 2.05 meters. **Hint Box:** - A hint box is included with an input box for entering an angle in degrees. To determine the angle of incidence \(\theta_1\), apply the concepts of refraction, taking into consideration Snell's Law and the geometry of the triangle formed by the light path. \[ \text{Snell's Law: } n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \] Where: - \(n_1\) is the refractive index of air (approx. 1.0) - \(n_2\) is the refractive index of water (approx. 1.33) Using these principles, solve for the angles using the provided diagram measurements.
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