A ray of light passes from air into water at an angle to the surface of the the water of 55°. Find the angle of refraction and the critical is the water
To find the angle of refraction, we can use Snell's Law, which 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 media:
n₁*sin(θ₁) = n₂*sin(θ₂)
Given:
- Angle of incidence (θ₁) = 55°
- Medium 1 (air) has a refractive index (n₁) approximately equal to 1.0003
- Medium 2 (water) has a refractive index (n₂) approximately equal to 1.333
We can rearrange Snell's Law to solve for θ₂:
sin(θ₂) = (n₁/n₂) * sin(θ₁)
Let's calculate the angle of refraction:
sin(θ₂) = (1.0003/1.333) * sin(55°)
θ₂ = sin^(-1)((1.0003/1.333) * sin(55°))
θ₂ ≈ 41.62°
Therefore, the angle of refraction is approximately 41.62°.
To find the critical angle, we can use the formula:
critical angle = sin^(-1)(1/n)
Where n is the refractive index ratio: n = n₂/n₁
Let's calculate the critical angle for water:
n = 1.333/1.0003
critical angle = sin^(-1)(1/n)
critical angle ≈ sin^(-1)(1/(1.333/1.0003))
critical angle ≈ sin^(-1)(0.7502)
critical angle ≈ 48.75°
Therefore, the critical angle for water is approximately 48.75°.
To find the angle of refraction, we can use Snell's Law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the speed of light in the first medium to the speed of light in the second medium.
Let's denote the angle of incidence as θ₁ and the angle of refraction as θ₂.
The speed of light in air is approximately equal to the speed of light in vacuum, so we can assume it to be the same for our calculation.
The speed of light in water is about 3/4 times the speed of light in air, so the ratio of the speed of light in the first medium to the speed of light in the second medium is 1/3/4 = 4/3.
Snell's Law can be written as follows:
sin(θ₁)/sin(θ₂) = 4/3
We are given that the angle of incidence is 55°, so let's substitute θ₁ = 55°:
sin(55°)/sin(θ₂) = 4/3
To find θ₂, we can rearrange the equation:
sin(θ₂) = sin(55°) * (3/4)
sin(θ₂) = (0.8192) * (3/4)
sin(θ₂) = 0.6144
To find the angle of refraction, we take the inverse sine of both sides:
θ₂ = sin^(-1)(0.6144)
θ₂ ≈ 38.7°
Therefore, the angle of refraction is approximately 38.7°.
The critical angle for water is the angle of incidence at which the refracted ray is parallel to the surface of the water. This occurs when the angle of refraction is 90°.
Using Snell's Law, we can find the critical angle by substituting θ₂ = 90° and solving for θ₁:
sin(θ₁)/sin(90°) = 4/3
sin(θ₁) = (4/3) * 1
sin(θ₁) = 4/3
To find the critical angle, we take the inverse sine of both sides:
θ₁ = sin^(-1)(4/3)
Using a calculator, we find that θ₁ ≈ 75.5°.
Therefore, the critical angle for water is approximately 75.5°.
To find the angle of refraction, we can apply Snell's law, which states that the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the speeds of light in the respective mediums. In this case, the mediums are air and water.
Snell's Law:
n₁sinθ₁ = n₂sinθ₂
Where:
n₁ = refractive index of air = 1 (since air is taken as the reference medium)
n₂ = refractive index of water = 1.33 (approximate value)
θ₁ = angle of incidence = 55° (given)
θ₂ = angle of refraction (to be found)
Rearranging the formula, we have:
sinθ₂ = (n₁ / n₂) * sinθ₁
Substituting the given values, we get:
sinθ₂ = (1 / 1.33) * sin(55°)
Using a scientific calculator, calculate the value of sinθ₂, which is approximately 0.619
θ₂ = sin^−1(0.619)
θ₂ ≈ 38.8°
Therefore, the angle of refraction is approximately 38.8°.
The critical angle is the angle of incidence for which the angle of refraction becomes 90°. This occurs when the light is going from a medium with a higher refractive index to one with a lower refractive index. In this case, the light is going from water (refractive index of 1.33) to air (refractive index of 1).
To find the critical angle, we can rearrange Snell's law as:
sinθ_c = (n₂ / n₁)
Substituting the values:
sinθ_c = (1.33 / 1)
Now, using a scientific calculator that has an inverse sine (sin^−1) function, find the value of sinθ_c, which is approximately 0.797.
θ_c = sin^−1(0.797)
θ_c ≈ 53.1°
Therefore, the critical angle for water is approximately 53.1°.