A light ray travels from water to air. If the ray strikes the water/air boundary at an angle of incidence of 46 degrees, what is the angle of refractiob in air? (n water = 1.33)
To determine the angle of refraction in air, you can use Snell's law, which relates the angle of incidence and the angle of refraction to the refractive indices of the two media. Snell's law is expressed as follows:
n1 * sin(theta1) = n2 * sin(theta2)
Where:
n1 = refractive index of the medium of incidence (water in this case)
n2 = refractive index of the medium of refraction (air in this case)
theta1 = angle of incidence
theta2 = angle of refraction
Let's plug in the values given in the question:
n1 = 1.33 (water's refractive index)
n2 = 1.00 (air's refractive index)
theta1 = 46 degrees (angle of incidence)
Using Snell's law, we can rearrange the equation to solve for theta2:
theta2 = arcsin((n1/n2) * sin(theta1))
Now substitute the values and solve:
theta2 = arcsin((1.33/1.00) * sin(46))
theta2 = arcsin(1.33 * sin(46))
Using a calculator, evaluate the expression:
theta2 ≈ 33.77 degrees
Therefore, the angle of refraction in air is approximately 33.77 degrees.