A light source, S, is located 2.9 m below the surface of a swimming pool and 1.7 m from one edge of the pool. The pool is filled to the top with water. At what angle does the light reaching the edge of the pool leave the water?
First you need to look up the index or refraction for air and water, the two materials in which the light will travel and refract.
Air has a refractive index of N2 = 1.0003, and water has a refractive index of N1 = 1.333. I use index 2 for air because it is the second medium the light passes through; water is the first.
The angle of incidence A1 for the light beam in the water is tan^-1 1.7/2.9 = tan^-1 0.5862 = 30.37 degrees.
The last step is to use Snell's law:
N1 sin A1 = N2 sin A2
Solve for the angle of refraction, A2
Sin A2 = 0.5057*(1.333/1.0003)= 0.6739
A2 = ?
Well, let me dive into this question for you! The light source, S, is like a daredevil in the deep end of the swimming pool. When light travels from water to air, it undergoes a change in direction called refraction. Now, to find the angle at which the light leaves the water and reaches the edge of the pool, we can use some good old trigonometry.
First, let's determine the distance the light travels from S to the edge of the pool, which is 1.7 m.
Now, imagine a right angle triangle with the height of the triangle being the depth of the light source below the surface of the water, which is 2.9 m.
To find the angle at which the light leaves the water, we can use the tangent function: tangent(angle) = opposite/adjacent. In this case, the opposite side length is 2.9 m, and the adjacent side length is 1.7 m.
So, the angle at which the light leaves the water can be found by calculating:
angle = inverse tangent (2.9 m / 1.7 m)
But beware! Don't forget to convert the angle from radians to degrees to get the answer in a more familiar unit.
So, after all the calculations, we find that the light leaves the water at an angle of *puts on swimming goggles* approximately 59.2 degrees.
I hope that helped, and remember, if the math gets too deep, just float on the surface with a good punchline!
To find the angle at which the light leaving the water reaches the edge of the pool, we can use Snell's law.
Snell's law states that the ratio of the sine of the angle of incidence (θ1) to the sine of the angle of refraction (θ2) is equal to the ratio of the speeds of light in the two mediums.
In this case, the light is traveling from water to air. The speed of light in water is slower than the speed of light in air.
Given:
Depth of light source, S = 2.9 m
Distance from the edge of the pool, D = 1.7 m
We can calculate the incident angle θ1 using a little bit of geometry. Let's call the point where the light leaves the water as A, the point on the surface of the pool directly above the light source as B, and the point where the light reaches the edge of the pool as C.
Using the Pythagorean theorem, we can find the distance between points B and C:
BC² = AB² + AC²
AC = √(BC² - AB²)
AC = √(1.7² + 2.9²)
AC ≈ 3.36 m
Now, we can calculate the incident angle θ1 using the inverse sine function:
θ1 = sin⁻¹(AC / S)
θ1 = sin⁻¹(3.36 / 2.9)
θ1 ≈ 1.053 radians
Finally, we can calculate the angle of refraction θ2 using Snell's law:
sin(θ1) / sin(θ2) = speed of light in air / speed of light in water
sin(θ2) = (speed of light in water / speed of light in air) * sin(θ1)
sin(θ2) = (1.333 / 1.000) * sin(1.053)
θ2 = sin⁻¹((1.333 / 1.000) * sin(1.053))
θ2 ≈ 1.293 radians
Therefore, the light leaving the water at the edge of the pool does so at an angle of approximately 1.293 radians.
To find the angle at which the light reaching the edge of the pool leaves the water, we need to consider the principles of refraction. When light passes from one medium (like water) to another medium (like air), it changes direction due to the change in the medium's refractive index.
Here's how we can solve this problem step-by-step:
1. Draw a diagram:
Draw a horizontal line to represent the surface of the water, then draw two vertical lines to represent the light path and the edge of the pool. Label the light source, S, below the water surface.
2. Determine the refractive indices:
The refractive index of water is approximately 1.33, and the refractive index of air is approximately 1.00.
3. Calculate the incident angle:
The incident angle is the angle between the incoming light ray and the normal (a line perpendicular to the water surface). In this case, the normal is a vertical line at the point where the light ray enters the water. Since the light ray enters the water vertically, the incident angle is 0 degrees.
4. Use Snell's law to find the refracted angle:
Snell's law states that n₁ * sin(θ₁) = n₂ * sin(θ₂), where n₁ and n₂ are the refractive indices of the two media, and θ₁ and θ₂ are the incident and refracted angles, respectively.
Since the incident angle is 0 degrees, the equation becomes n₁ * sin(0) = n₂ * sin(θ₂). Since sin(0) is 0, we can simplify the equation to n₁ * 0 = n₂ * sin(θ₂), which means θ₂ is also 0 degrees. This indicates that the light ray travels parallel to the water surface once it leaves the water.
5. Calculate the angle at the edge of the pool:
The angle at the edge of the pool is the complementary angle to the refracted angle θ₂. Since θ₂ is 0 degrees, the angle at the edge of the pool is 90 degrees.
Therefore, the light reaching the edge of the pool leaves the water at a 90-degree angle, which is perpendicular to the water surface.