A beam of light is emitted in a pool of water from a depth of 64.0 cm. How far away, relative to the spot directly above it, must it strike the air-water interface in order that the light does not exit the water?
theta 2 must be at least 90 degrees so 1*sin90=1.333sin(theta2) theta2 = 48.61 degrees. if distance from spot above the light to this point = x then (x/64)=tan48.61 x= 72.619 cm
Well, if we're talking about light not wanting to leave the water, it seems like the light is really committed to being a "pool" player. Haha! But let's dive into the question.
To solve this, we can use Snell's Law, which relates the angles and the indices of refraction of the two media. In this case, we have air (n₁ = 1) and water (n₂ = 1.33). The formula for Snell's Law is:
n₁*sin(θ₁) = n₂*sin(θ₂)
Now, the angle θ₁ depends on how far the light strikes the interface relative to the spot directly above it. So, to answer your question, we need to find the value of θ₁ that would cause the light to not exit the water.
But hey, don't worry, I've got a "watery" good answer for you! Since the question doesn't mention any other angles or specific directions, we can assume the light strikes the interface head-on, perpendicular to it. In that case, θ₁ would be 90 degrees.
So, substituting the values into Snell's Law:
1 * sin(90) = 1.33 * sin(θ₂)
sin(90) is equal to 1, making the equation:
1 = 1.33 * sin(θ₂)
Now, we need to find the angle θ₂ that will cause sin(θ₂) to be equal to or greater than 1. This will result in the light being totally reflected back into the water and not exiting.
However, according to the laws of physics, the maximum value of sin(θ₂) is 1. So, sin(θ₂) can't be equal to or greater than 1. Therefore, the light will always exit the water, no matter how far away it strikes the air-water interface.
So, it looks like this light beam is determined to escape the watery depths! Our light is a "free diver". Haha!
To determine the distance at which the light beam must strike the air-water interface so that it does not exit the water, we can use Snell's law. Snell's law relates the angle of incidence of a ray of light with the refractive indices of the two media it is traveling through.
Here are the steps to find the distance:
Step 1: Identify the given values:
- Depth of the beam of light in water (h): 64.0 cm
- Refractive index of water (n₁): 1.33 (approximately)
Step 2: Convert the depth from centimeters to meters:
- Depth in meters (h): 64.0 cm = 64.0 cm / 100 cm/m = 0.64 m
Step 3: Determine the refractive index of air (n₂):
- Refractive index of air (n₂): 1.00 (approximately)
Step 4: Apply Snell's law:
The formula for Snell's law is:
n₁ * sin(θ₁) = n₂ * sin(θ₂)
In this case, we want the light to be totally reflected at the air-water interface, which means it should not exit the water. This occurs when the angle of refraction (θ₂) is 90 degrees, making sin(θ₂) equal to 1.
Substituting the known values into the equation, we have:
1.33 * sin(θ₁) = 1 * 1
Step 5: Solve for sin(θ₁):
sin(θ₁) = 1 / 1.33
Step 6: Find the angle of incidence (θ₁) using the inverse sine function:
θ₁ = arcsin(1 / 1.33)
Step 7: Calculate the distance from the spot directly above to where the beam of light must strike the air-water interface:
s = h * tan(θ₁)
Substituting the known values into the equation, we have:
s = 0.64 m * tan(arcsin(1 / 1.33))
Step 8: Calculate the distance s using a calculator:
Use a scientific calculator to find the value of tan(arcsin(1 / 1.33)) and multiply it by 0.64 m to get the final value.
The result will give you the distance, relative to the spot directly above, at which the light beam must strike the air-water interface so that it does not exit the water.
To solve this problem, we need to use the principles of refraction and Snell's law. Snell's law states that the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the velocities of light in the two media.
Let's assume the speed of light in air is approximately equal to the speed of light in vacuum, which is approximately 3.0 x 10^8 meters per second.
Given:
Depth of water (h) = 64.0 cm = 0.64 m
Speed of light in air (v₁) = 3.0 x 10^8 m/s
We need to find the distance from the spot directly above the light source to the point where the light strikes the air-water interface without exiting the water.
To find this distance (let's call it x), we can use Snell's law:
n₁ * sin(θ₁) = n₂ * sin(θ₂)
Since the light is leaving the water, the angle of refraction in the water (θ₂) will be 90 degrees.
n₁ * sin(θ₁) = n₂ * sin(90)
The refractive index of air is very close to 1, so we can simplify the equation:
sin(θ₁) = n / 1
sin(θ₁) = n
Now, we need to find the refractive index of water. The refractive index (n) is the ratio of the speed of light in vacuum to the speed of light in a given medium. It can be calculated as follows:
n = v₁ / v
where v₁ is the speed of light in air, and v is the speed of light in water.
To find the speed of light in water, we can use its relation to the speed of light in a vacuum:
v = c / n
where c is the speed of light in a vacuum, and n is the refractive index of water.
Now, we can calculate the refractive index of water:
n = v₁ / v
Substituting the values:
n = (3.0 x 10^8 m/s) / (2.25 x 10^8 m/s) (the refractive index of water is approximately 1.33)
Now, we can find the angle of incidence (θ₁) using the sine function:
sin(θ₁) = n
θ₁ = arcsin(n)
Substituting the value of the refractive index of water:
θ₁ = arcsin(1.33)
Using a calculator, we find that θ₁ ≈ 51.54 degrees.
Now, we need to find the distance (x) from the spot directly above the light source to the point where the light strikes the air-water interface. We can use the tangent function:
tan(θ₁) = h / x
Substituting the value of θ₁ and h:
tan(51.54) = 0.64 / x
Rearranging the equation to solve for x:
x = 0.64 / tan(51.54)
Using a calculator, we find that x ≈ 0.817 m (rounded to three decimal places).
Therefore, the light must strike the air-water interface approximately 0.817 meters away from the spot directly above it in order not to exit the water.