Light passes from air into water at an angle of incidence of 40.4°. Determine the angle of refraction in the water.
I hope you have heard of Snell's law.
N(air)*sin theta(in air)
= N(water)*sin theta (in water)
sin theta (in water) = (1/1.33)sin 40.4 = 0.487
Solve for theta in water
To determine the angle of refraction in water, 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 constant and equal to the ratio of the speeds of light in the two media:
n1 * sin(θ1) = n2 * sin(θ2)
Where:
n1 = refractive index of the initial medium (air) = 1 (approximately)
n2 = refractive index of the second medium (water) = 1.33 (approximately)
Given that the angle of incidence (θ1) is equal to 40.4°, we can substitute the values into Snell's law to find the angle of refraction (θ2):
1 * sin(40.4°) = 1.33 * sin(θ2)
Rearranging the equation to solve for θ2:
sin(θ2) = (1 * sin(40.4°)) / 1.33
Using a scientific calculator, calculate the value inside the parentheses on the right side:
sin(θ2) = 0.6075 / 1.33
Then, find the arcsine of the result to obtain the angle of refraction, θ2:
θ2 = arcsin(0.6075 / 1.33)
Using a scientific calculator, calculate the arcsine (inverse sine) of 0.6075 divided by 1.33:
θ2 ≈ 46.3°
Therefore, the angle of refraction in water is approximately 46.3°.