A beam of light in air is incident on a stack of four flat...

A beam of light in air is incident on a stack of four flat transparent materials with indices of refraction 1.20 $1.40,1.32,$ and $1.28 .$ If the angle of incidence for the beam on the first of the four materials is $60.0^{\circ},$ what angle does the beam make with the normal when it emerges into the air after passing through the entire stack?

Key Concepts

Plane-Parallel Interfaces

When light passes through a stack of materials with plane-parallel interfaces, the cumulative effect of refraction at each boundary can be analyzed using Snell's Law at every interface. For a set of parallel layers, despite individual changes in direction, the emerging beam in the original medium will have the same angle relative to the normal as the incident beam due to the symmetry and reversibility of the light path.

Refractive Index

The refractive index is a measure of how much the speed of light is reduced in a medium compared to its speed in a vacuum. It is a central concept in understanding refraction, as differences in refractive indices between materials determine the degree to which light bends at an interface.

Refraction

Refraction is the bending of light as it passes from one medium to another with a different optical density. The change in speed causes the light to change direction at the boundary between the media, which is critical when analyzing how light propagates through layered materials.

Snell's Law

Snell's Law quantitatively describes refraction by relating the incident angle and the refracted angle to the indices of refraction of the two media. It is expressed as n? sin?? = n? sin??, and it allows us to calculate the change in the angle of the light as it passes from one material to another.

Transcript

00:01 Okay, in this problem, you have stacked materials that a ray is traveling through, and each one of these materials has a different index of refraction, and they want to know what is the angle at which this ray exits these materials.

00:25 So for each of these transitions, we're going to use snell's law.

00:30 To get started we'll set it up n1 and all of these values were given to you in the problem n2 sine theta 2 and i'll do a couple of these examples for you you plug in what you know 1 .0 sine 60 and it's going into this material with an index of refraction of 1 .2 and we want to solve for the angle at which it is refracted.

01:06 So to solve for that, we need to get theta 2 alone.

01:12 And whenever we rearrange this equation, we get the inverse of sign because in order to get this unattached from that angle, you do the inverse.

01:24 So sine, inverse sign.

01:27 And it would be 1 .0.

01:31 0 .60 all over 1 .2.

01:39 And when you calculate that, you can angle around 46 .19 degrees.

01:50 And we're going to use that for the transition between n2 and n3 or material 2 and material 3.

02:01 So we just solved for theta 2.

02:08 So now we're going to do n2 sine theta 2 equal to n3, sine theta 3.

02:20 Plug in what you know here.

02:21 So 1 .2, sine 46 .19, equal to 1 .4, sine theta 3.

02:37 And we're going to solve for theta 3, which is the inverse sign of 1 .2, sign 46 .19 divided by 1 .4.

02:58 It gives us an angle of 38 .21 degrees.

03:09 And we'll keep going here...