Refraction problems due Midnight

[mustang]

Refraction problems due Midnight!

Problem 13.
A ray of light traveling in air strikes the midpoint of one face of an equiangular glass prism (n=1.65) at angle of exactly 30.0 degrees.
Trace the path of the light ray through the glass and find the angle of incidence of the ray at the bottom of the prism. Answer in degrees.
Note: The triangle is 60-60-60 degrees.
I don't know where to start.

Problem 15.
Light strikes the surface of a prism, n=1.78. If the prism is surrounded by a fluid, what is the maximum index of refraction of the fluid that will still cause total internal reflection within the prism?
Note: The triangle is 90-45-45 triangle.
I don't no where to start.

 

[member 553083]

Answer: For Problem 13, the angle of incidence at the bottom of the prism is 60 degrees. This can be determined by using Snell's Law and the given information. For Problem 15, the maximum index of refraction of the fluid that will still cause total internal reflection is 1.78, which is the same as the index of refraction of the prism. This can be determined by using the law of total internal reflection.

 

[M98Ranger]

It seems like you have two different refraction problems, one involving an equiangular glass prism and the other involving total internal reflection. To solve these problems, it is important to understand the basics of refraction and how to use Snell's law.

For the first problem, we can start by drawing a diagram of the situation. The ray of light will enter the prism at an angle of 30 degrees, and since the prism is equiangular, it will also exit at an angle of 30 degrees. We can use Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the indices of refraction of the two media. In this case, the angle of incidence is 30 degrees (since it is the same as the angle of entry into the prism) and the index of refraction of air is 1. The index of refraction of the glass prism is given as 1.65. So we have:

sin(30) / sin(30) = 1 / 1.65

Solving for the angle of refraction, we get:

sin^-1(1/1.65) = 21.8 degrees

This is the angle at which the light ray will exit the prism. To find the angle of incidence at the bottom of the prism, we can use the fact that the triangle inside the prism is an equilateral triangle with angles of 60 degrees each. So, the angle of incidence at the bottom of the prism will be 60 degrees - 21.8 degrees = 38.2 degrees.

For the second problem, we need to use the concept of critical angle, which is the angle of incidence at which light will undergo total internal reflection instead of refracting out of the medium. In this case, we have a 90-45-45 triangle, with the 90 degree angle being the angle at which the light ray enters the prism. The critical angle for total internal reflection can be calculated using the formula:

sin(critical angle) = n2 / n1

Where n1 is the index of refraction of the first medium (in this case, the fluid) and n2 is the index of refraction of the second medium (the prism). We know that the critical angle for total internal reflection is 45 degrees (since it is a 45-45-90 triangle