A horizontal light ray is incident on a prism as shown in the figure with Problem 18 where βis a

A horizontal light ray is incident on a prism as shown in...

Key Concepts

Snell's Law

Snell's Law is a fundamental principle in optics that relates the angles of incidence and refraction when light passes from one medium into another with a different index of refraction. It states that n? sin(??) = n? sin(??), which is essential for understanding how light bends upon entering or leaving a prism.

Prism Geometry

Prism geometry concerns the shape and angles within a prism, which determine how light is refracted at each interface. The specific layout of the prism, particularly the relationship between its angles and faces, plays a critical role in calculating the net deviation of light as it passes through.

Small Angle Approximation

The small angle approximation is a mathematical simplification used when dealing with very small angles, where sin(?) is approximately equal to ? (when measured in radians). This approximation simplifies the trigonometric relationships in optical calculations, making it easier to show proportional relationships like the deviation angle being proportional to the small angle ?.

Angle of Deviation

The angle of deviation represents the overall change in direction of a light ray after it passes through a prism or another refractive medium. It is defined as the difference between the angle of the incident ray and the angle at which the ray emerges, and is a key quantity in understanding the optical behavior of prisms.

Transcript

00:01 For this problem, we want to consider a light ray incident on a prism.

00:08 And we want to solve for the angle delta, which is the angle between the incident ray of light and the ray of light that exits the prism on the other side in terms of the indexes of refraction and the angles of the prism.

00:27 In order to do this, i'm going to first go step by.

00:31 Step and try to identify all of the angles that are involved with these two rays, starting with the angle of incidence for the incident ray.

00:43 So i'm going to zoom in here on where our light ray is entering the prism.

00:50 We can draw the line that is perpendicular to the surface of the prism.

00:58 And we're interested here in theta 1, the angle of incidence.

01:04 We know because of similar triangles that this angle here is going to be alpha, which means that this angle here will be alpha.

01:12 And so from this we can pretty easily figure out that theta 1 is going to be 90 degrees minus alpha because theta 1 plus alpha has to be equal to 90 degrees based off of the figure that we've drawn.

01:32 Now we will want to consider the ray that is refracted as it moves into the prism, and we want to find theta to the angle of refraction.

01:44 We can do this using snell's law.

01:48 So outside of the prism, we have an index of refraction that i'm going to label as n1.

01:54 Inside the prism, it tells us the index of refraction is n.

01:59 And so we have n1.

02:03 Sine theta 1 equals n sine theta 2.

02:11 So if we are solving here for theta 2, you would get that theta 2 is equal to inverse sine of n1 over n times sign of theta 1.

02:32 And we already know what theta 1 is equal to, so i'm not going to plug that in just yet in order to save us a little bit of space here.

02:45 Now we want to figure out what will be the angle of incidence for the ray as it hits the other side of the prism.

02:55 So again, we're going to zoom in on the other side of the prism.

03:00 We've got a light ray here that is exiting the prism.

03:07 And we want to figure out what is the angle.

03:12 I'm going to label it theta 3 here of it as it exits.

03:18 In order to solve this, i'm going to go ahead and add in the other side of the prism.

03:31 Because we now have a triangle that will be really helpful in solving.

03:37 Because we know that this angle on the right hand side is going to be 9 .5.

03:43 90 minus theta 3.

03:46 And that this angle right here on the left hand side will be 90 minus theta 2.

03:52 Because, again, theta 2, if we go from theta 2 and then add in this angle, we get a 90 degree angle.

04:03 Same on the right hand side.

04:05 So now we have a triangle.

04:07 We know all three angles in that triangle.

04:09 So we can go ahead and set up the equation that beta.

04:15 Plus 90 minus theta 2 plus 90 minus theta 3 is all equal to 180 degrees.

04:27 And of course, now we want to solve for the angle theta 3.

04:32 When you solve for theta 3, you would find that theta 3 is equal to beta minus theta 2.

04:39 And again, we know what theta 2 is in terms of theta 1, and we know what theta 1 is in terms of the angles of the prism.

04:47 So we can move right along.

04:50 And the last angle we need to figure out here is the angle of refraction.

04:58 I'm going to label this theta 4 as the light wave exits the prism.

05:05 And we're going to use snell's law to do that again because we know theta 3.

05:10 And we're trying to find the angle of refraction theta 4.

05:14 So in this case, we have n sine theta 3 will equal n1, sine theta 4...

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