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To solve this problem, we need to demonstrate that Snell's Law, which is given by:

\[
\frac{\sin \theta_1}{\sin \theta_2} = \frac{v_1}{v_2}
\]

is satisfied when a ray of light travels from point \( A \) in air to point \( B \) in water along the path \( ACB \), such that the total time taken is minimized. Here, \( v_1 \) is the velocity of light in air, \( v_2 \) is the velocity of light in water, \( \theta_1 \) is the angle of incidence, and \( \theta_2 \) is the angle of refraction.

### Step-by-Step Solution:

1. **Define the distances and times:**

   - Let the point \( C \) be the point on the water surface where the light ray refracts.
   - Let the distance from \( A \) to \( C \) (horizontal) be \( x \) and from \( C \) to \( B \) (horizontal) be \( d - x \), where \( d \) is the total horizontal distance between \( A \) and \( B \).
   - Let the vertical height from \( A \) to the water surface be \( h_1 \) and from the surface to \( B \) be \( h_2 \).

2. **Express the travel times in each medium:**

   - The time taken for the light to travel from \( A \) to \( C \) in air is:
   \[
   t_1 = \frac{\sqrt{x^2 + h_1^2}}{v_1}
   \]
   
   - The time taken for the light to travel from \( C \) to \( B \) in water is:
   \[
   t_2 = \frac{\sqrt{(d - x)^2 + h_2^2}}{v_2}
   \]

3. **Total time taken, \( T \):**

   The total time \( T \) is the sum of \( t_1 \) and \( t_2 \):

   \[
   T(x) = \frac{\sqrt{x^2 + h_1^2}}{v_1} + \frac{\sqrt{(d - x)^2 + h_2^2}}{v_2}
   \]

4. **Minimize the total time \( T(x) \):**

   To minimize \( T(x) \), we differentiate \( T(x) \) with respect to \( x \) and set the derivative equal to zero:

   \[
   \frac{dT}{dx} = \frac{d}{dx} \left( \frac{\sqrt{x^2 + h_1^2}}{v_1} + \frac{\sqrt{(d - x)^2 + h_2^2}}{v_2} \right) = 0
   \]

5. **Apply the derivative:**

   Differentiate each term separately:

   \[
   \frac{dT}{dx} = \frac{x}{v_1 \sqrt{x^2 + h_1^2}} - \frac{(d - x)}{v_2 \sqrt{(d - x)^2 + h_2^2}} = 0
   \]

6. **Rearrange the equation:**

   Rearrange the equation to find:

   \[
   \frac{x}{\sqrt{x^2 + h_1^2}} = \frac{v_1}{v_2} \cdot \frac{(d - x)}{\sqrt{(d - x)^2 + h_2^2}}
   \]

7. **Use the definitions of trigonometric functions:**

   Notice that:

   \[
   \sin \theta_1 = \frac{x}{\sqrt{x^2 + h_1^2}}, \quad \sin \theta_2 = \frac{(d - x)}{\sqrt{(d - x)^2 + h_2^2}}
   \]

   Substituting these into the equation gives:

   \[
   \frac{\sin \theta_1}{\sin \theta_2} = \frac{v_1}{v_2}
   \]

8. **Conclusion:**

   This is Snell's Law, which confirms that the path taken by the light ray satisfies the principle of least time, as required.

Would you like a further breakdown of any step, or have any other questions? Here are some related topics you might be interested in:

1. How to apply Snell's Law in different mediums?
2. The derivation of Snell's Law using Huygens' Principle.
3. Analyzing the refraction at the boundary of different shapes (e.g., spherical surfaces).
4. Understanding the connection between Snell's Law and the wavefront approach.
5. Practical applications of Snell's Law in optics.

**Tip:** Always remember that Snell's Law can be derived from Fermat's Principle, which states that the path taken by light between two points is the one that minimizes the travel time.

Let v1 be the velocity of light in air and v2 the velocity of light in water. According to Fermat's Principle, a ray of light will travel from a point A in the air to a point B in the water by a path ACB that minimizes the time taken. Show that sin(theta1)/sin(theta2) = v1/v2 where theta1 (the angle of incidence) and theta2 (the angle of refraction) are as shown. This equation is known as Snell's Law.

Learn how to derive Snell's Law from Fermat's Principle. This problem involves the path of light traveling from air to water and covers key concepts in optics and refraction, suitable for Grades 11-12.

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