The 45-45-90 triangular prism shown in the figure has an index of refraction of 1.58, and is surrounded by air. A ray of light is incident on the left face at an angle of 51.0°. The point of incidence is high enough that the refracted ray hits the opposite sloping side.
1. Through which face of the prism does the light exit? left right, bottom.
2. Calculate the angle at which the ray exits the prism, relative to the normal of that surface. Express your answer as a negative number if the angle is clockwise relative to the normal, and a positive number for a counterclockwise angle
3.What is the maximum index of refraction for which the ray exits the prism from the right-hand side? (It may help to remember that sin(90° - è) = cos è.)
To determine the answers to these questions, we need to use Snell's Law, which relates the angles of incidence and refraction when a ray of light passes through a boundary between two mediums.
1. To determine through which face of the prism the light exits, we need to assess the relationship between the angles of incidence and refraction. Snell's Law 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 mediums:
n1 * sin(θ1) = n2 * sin(θ2)
where n1 and n2 are the indices of refraction of the initial and final mediums, θ1 is the angle of incidence, and θ2 is the angle of refraction.
In this case, the light is incident on the left face at an angle of 51.0°. Since the point of incidence is high enough that the refracted ray hits the opposite sloping side, we can conclude that the light will exit through the bottom face of the prism.
Therefore, the answer to question 1 is: bottom.
2. To calculate the angle at which the ray exits the prism relative to the normal of that surface, we need to apply Snell's Law again. However, we first need to calculate the angle of refraction at the left face.
Using Snell's Law: n1 * sin(θ1) = n2 * sin(θ2), where n1 is the index of refraction of air (1.00) and n2 is the index of refraction of the prism (1.58), and θ1 is the angle of incidence (51.0°), we can solve for θ2:
1.00 * sin(51.0°) = 1.58 * sin(θ2)
Now, calculate θ2:
θ2 = arcsin((1.00 * sin(51.0°)) / 1.58)
The resulting value of θ2 will be the angle at which the ray exits the prism relative to the normal of the bottom surface.
Therefore, the answer to question 2 is the value of θ2 calculated above.
3. To find the maximum index of refraction for which the ray exits the prism from the right-hand side, we need to determine the critical angle, beyond which total internal reflection occurs.
The critical angle can be found using the formula: sin(θc) = 1 / n, where n is the ratio of the indices of refraction (n1/n2). In this case, n1 is the index of refraction of air (1.00) and n2 is the maximum index of refraction we want to find.
Rearrange the formula to solve for n2:
n2 = 1 / sin(θc)
Since the ray exits the prism from the right-hand side, the incident angle at the right face would be the angle of incidence at the left face (51.0°). Therefore, we can find θc using the formula mentioned in question 2:
θc = arcsin((1.00 * sin(51.0°)) / 1.58)
Finally, calculate n2:
n2 = 1 / sin(θc)
The resulting value of n2 will be the maximum index of refraction for which the ray exits the prism from the right-hand side.
Therefore, the answer to question 3 is the value of n2 calculated above.