Please help me!!!!
A swimmer looks up from underwater in a pool and sees a tree. The light traveling in the air makes an ancle of 40 degrees, with the surface of the water. Calculate the anglethe light ray makes with the surface of the water ( n= 1.33) while traveling in the water
i know the answer is 55 degrees, but i have NO idea how to get it :(
Use Snell's law, which states that
n1 * sin theta1 = n2 * sin theta2
The "theta" angles are the angles the ray makes with the perpendicualr, not the surface. The n's are the indexes of refraction in the two differient media.
For air, the angle is theta2 = 50 degrees and n2 = 1.00
For water, the angle is theta1 and n = 1.33
So, solve this equation:
1.33 sin theta1 = 1 * sin 50
sin theta1 = 0.576
theta1 = 35.2 degrees
The angle of the ray with the surface is 90 degrees minus that, or 54.8 degrees. They want you to round that to 55 degrees
Ignore Natosha's incorrect answer
alright, so does anybody have a right answer? lol
Trust drwls's answer.
To determine the angle that the light ray makes with the surface of the water, we can apply the laws of refraction.
First, let's define the given angles:
θ1 = 40 degrees (angle of incidence in air)
θ2 = ? (angle of refraction in water)
The relationship between the angles of incidence and refraction when light passes from one medium to another is given by Snell's law:
n1 * sin(θ1) = n2 * sin(θ2),
where n1 is the refractive index of the incident medium (air) and n2 is the refractive index of the refracted medium (water).
Given that n1 = 1 (since air's refractive index is approximately 1) and n2 = 1.33 (as provided in the question), we can rewrite Snell's law as:
1 * sin(θ1) = 1.33 * sin(θ2).
Now, substitute the given value for θ1 (40 degrees) and solve for θ2:
sin(40) = 1.33 * sin(θ2).
To isolate sin(θ2), divide both sides of the equation by 1.33:
sin(θ2) = sin(40) / 1.33.
Take the inverse sine (sin^(-1)) of both sides to find θ2:
θ2 ≈ sin^(-1)(0.642) ≈ 40.92 degrees.
Therefore, the angle that the light ray makes with the surface of the water is approximately 40.92 degrees.
It seems there might be a discrepancy between the given answer (55 degrees) and the calculated value (40.92 degrees). Please double-check the provided information or equations to verify the desired angle.