What Is the Minimum Angle of Incidence for Light to Traverse a 60-Degree Prism?

[John_Galt]

Homework Statement

Given an apex angle of 60 degrees and an index of refraction of 1.6; what is the smallest angle of incidence for which a ray can enter the left face of a prism and exit the right face?

Homework Equations

The Attempt at a Solution

 

[Chi Meson]

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https://www.physicsforums.com/showthread.php?t=94379

 

[Moetasim]

The smallest angle of incidence for which a ray can enter the left face of a prism and exit the right face can be found using the 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 incident medium is air with an index of refraction of 1 and the refractive medium is the prism with an index of refraction of 1.6. Therefore, the equation can be written as sin(angle of incidence)/sin(angle of refraction) = 1/1.6.

Since the apex angle of the prism is 60 degrees, the angle of refraction will also be 60 degrees. Solving for the angle of incidence, we get sin(angle of incidence) = (1/1.6)sin(60) = 0.625. Taking the inverse sine, we get an angle of incidence of approximately 38.2 degrees.

Therefore, the smallest angle of incidence for which a ray can enter the left face of a prism and exit the right face is approximately 38.2 degrees. It is important to note that this is the minimum angle of incidence and any angle larger than this will also allow the ray to enter and exit the prism.