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?
To find the angle of the light ray when it enters the water, we need to use the concept of Snell's Law, which relates the angles of incidence and refraction with the refractive indices of the two media.
Snell's Law states:
n1 * sin(theta1) = n2 * sin(theta2)
Where:
- n1 is the refractive index of the medium the light ray is coming from (outside the aquarium)
- n2 is the refractive index of the medium the light ray is entering (water)
- theta1 is the angle of incidence
- theta2 is the angle of refraction
First, let's find the refracted angle when the light ray enters the glass. Given that the angle of incidence is 43.5 degrees and the refractive index of the glass is 1.48, we can rearrange Snell's Law to solve for theta2:
sin(theta2) = (n1/n2) * sin(theta1)
sin(theta2) = (1/1.48) * sin(43.5)
Using a scientific calculator or trigonometric tables, calculate sin(theta2).
sin(theta2) ≈ 0.5603
Now, to find the angle of the light ray when it enters the water, we need to apply Snell's Law again, using the refractive index of water (assume 1.33) and the calculated value of sin(theta2) as follows:
n1 * sin(theta1) = n2 * sin(theta2)
(1.48) * sin(27.7) = (1.33) * sin(theta2)
Rearrange the equation and solve for theta2:
sin(theta2) = (1.48/1.33) * sin(27.7)
sin(theta2) ≈ 0.7375
Use a scientific calculator or trigonometric tables to find the inverse sine (sin^-1) of 0.7375 to get the angle theta2:
theta2 ≈ sin^-1(0.7375)
theta2 ≈ 47.52 degrees
Therefore, the angle of the light ray when it enters the water is approximately 47.52 degrees.
To find the angle of the light ray when it enters the water, we need to apply Snell's law, which relates the angles and refractive indices of the incident and refracted rays.
The formula for Snell's law is: n1 * Sin(theta1) = n2 * Sin(theta2),
Where:
- n1 and n2 are the refractive indices of the mediums (air and water, glass and water in this case).
- theta1 is the angle of incidence (the angle between the incident ray and the perpendicular).
- theta2 is the angle of refraction (the angle between the refracted ray and the perpendicular).
Given:
- n1 (glass) = 1.48
- theta1 (incident angle) = 43.5º
- theta2 (refracted angle in the glass) = 27.7º
We need to find theta2 (refracted angle in water).
First, let's calculate the refractive index of water (n2) using the refractive index of glass (n1) and Snell's law:
n1 * Sin(theta1) = n2 * Sin(theta2)
1.48 * Sin(43.5º) = n2 * Sin(27.7º)
0.9886 = n2 * 0.4682
n2 = 0.9886 / 0.4682
n2 ≈ 2.11
Now we can use Snell's law again, but this time with the refractive index of water (n2) to find the angle of the light ray when it enters the water (theta3).
n1 * Sin(theta2) = n2 * Sin(theta3)
1.48 * Sin(27.7º) = 2.11 * Sin(theta3)
0.6937 = 2.11 * Sin(theta3)
Sin(theta3) = 0.6937 / 2.11
Sin(theta3) ≈ 0.3287
To find theta3, we take the inverse Sin of 0.3287:
theta3 ≈ Sin^(-1)(0.3287)
theta3 ≈ 19.2º
Therefore, the angle of the light ray when it enters the water is approximately 19.2º.