An aquarium filled with water has flat glass sides whose index of refraction is 1.48. A beam of light from outside the aquarium strikes the glass at a 43.5º angle to the perpendicular . The angle this light ray enters the glass is 27.7º, what is the angle of this light ray when it enters the water?
The angle in the glass ( r ) related to the angle of incidence (i) as
sin i/sin r = n2/n1,
sin r = n1•sin i/n2 = 1•sin 43.5/1.48 = =0.465.
r = arcsin 0.465= 27.7degr.
Because the surfaces are parallel, the refraction angle from the first surface is the incident angle at the second surface.
sin r/sin φ =n3/n2,
sin φ = n2•sin r/n3 =1.48•0.465/1.33 =0.517,
φ = arcsin 0.517 =31.16 degr.
Thank you, I did that but I did not round until the end so I had 31.06º which was listed as incorrect
To find the angle of the light ray when it enters the water, we can use the law of refraction (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 indices of refraction of the two media.
The formula for Snell's law is:
n1 * sin(θ1) = n2 * sin(θ2)
Where:
n1 = index of refraction of the first medium (air)
θ1 = angle of incidence (from outside the aquarium)
n2 = index of refraction of the second medium (glass)
θ2 = angle of refraction (inside the glass)
In this case, we are given:
n1 (air) = 1 (approximation)
θ1 = 43.5º
n2 (glass) = 1.48 (index of refraction of the glass)
To find θ2, we can rearrange the equation and solve for θ2:
sin(θ2) = (n1 / n2) * sin(θ1)
θ2 = sin^(-1)((n1 / n2) * sin(θ1))
Substituting the given values:
θ2 = sin^(-1)((1 / 1.48) * sin(43.5º))
θ2 ≈ 29.61º
Now, to find the angle when the light ray enters the water, we need to apply Snell's law again. In this case, the indices of refraction are:
n1 (glass) = 1.48
θ1 (inside the glass) = 27.7º
n2 (water) = 1.33 (index of refraction of water)
Using the same formula as before:
sin(θ2) = (n1 / n2) * sin(θ1)
θ2 = sin^(-1)((n1 / n2) * sin(θ1))
Substituting the given values:
θ2 = sin^(-1)((1.48 / 1.33) * sin(27.7º))
θ2 ≈ 34.58º
Therefore, the angle of the light ray when it enters the water is approximately 34.58º.
To find the angle of the light ray when it enters the water, we can use the laws of refraction. The law of refraction, also known as Snell's law, relates the angles of incidence and refraction to the indices of refraction of the materials involved.
Snell's law states that:
n1 * sin(theta1) = n2 * sin(theta2)
where:
- n1 is the index of refraction of the first material (in this case, glass)
- theta1 is the angle of incidence of the light ray in the first material
- n2 is the index of refraction of the second material (in this case, water)
- theta2 is the angle of refraction of the light ray in the second material
Given:
- The index of refraction of the glass, n1, is 1.48
- The angle of incidence, theta1, is 27.7º
- We need to find the angle of refraction, theta2, in water
First, we need to find the index of refraction of water, n2. The index of refraction of water is approximately 1.33.
Now, we can rearrange Snell's law to solve for theta2:
n2 * sin(theta2) = n1 * sin(theta1)
Substituting the given values:
1.33 * sin(theta2) = 1.48 * sin(27.7º)
Now, we can solve this equation for sin(theta2):
sin(theta2) = (1.48 * sin(27.7º)) / 1.33
Using a calculator, we find that sin(theta2) is approximately 0.492.
To find the angle of refraction, theta2, we need to take the arcsine (sin^-1) of 0.492:
theta2 = sin^-1(0.492)
Using a calculator, we find that theta2 is approximately 29.6º.
Therefore, the angle of this light ray when it enters the water is approximately 29.6º.