In Figure 26.6, suppose that the angle of incidence is θ1=30.0^∘ , the thickness of the glass pa

In Figure 26.6, suppose that the angle of incidence is...

In Figure $26.6,$ suppose that the angle of incidence is $\theta_{1}=30.0^{\circ}$ , the thickness of the glass pane is 6.00 $\mathrm{mm}$ , and the refractive index of the glass is $n_{2}=1.52 .$ Find the amount (in mm) by which the emergent ray is displaced relative to the incident ray.

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 between two media with different refractive indices. It is expressed by the formula n?sin?? = n?sin??, where n denotes the refractive index of each medium and ? represents the angle of the light ray relative to the normal. This law is essential for predicting how light will bend or change direction during refraction.

Refractive Index

The refractive index quantifies how much light slows down when it travels through a medium compared to its speed in a vacuum. It is a key property of optical materials and determines the degree of bending of light at an interface between different media. Different refractive indices between materials cause the light to change direction according to Snell's Law.

Lateral Displacement in a Slab

When a light ray passes through a medium like a glass slab with parallel faces, it undergoes refraction twice—once when entering and once when exiting the slab. Although the emergent ray remains parallel to the incident ray, the cumulative effect of these refractions results in a lateral shift, meaning the ray is displaced sideways. The amount of lateral displacement depends on the thickness of the slab, the angles of incidence and refraction, and the refractive index of the material.

Geometric Optics of Plane-Parallel Interfaces

In optical systems involving flat, parallel surfaces, such as a glass pane, the principles of geometric optics are applied to trace the paths of light rays through the media. This involves using the laws of reflection and refraction to determine the exact trajectory and displacement of the light ray as it travels through the slab, accounting for factors like the thickness of the pane and the angles at which the light enters and exits.

Transcript

00:01 So we have a glass slab and light comes in through here at an angle of 30 degrees.

00:11 And n1 for air is 1.

00:14 And n2 for glass is 1 .523.

00:20 And first thing we're going to do is we're going to find where would this light go and strike if it was not refracted.

00:27 So if this angle is 30 degrees, then this angle is also 30 degrees.

00:32 And we are trying to find this distance x, given that the thickness of the glass slab is 6 millimeters.

00:40 So in this right triangle, 6 is the adjacent, and x is the opposite.

00:46 So we have tangent of 30 degrees is equal to x over 6, and that gives me x is equal to 3 .46 millimeters.

00:58 So the next thing we need to figure out is where does this light actually go? in strike.

01:03 Now because of the refractive index of glass light will be deflected and it would actually go and hit some point and that point would be at the distance x -prime.

01:15 So next thing we want to do is we want to figure out what is x -prime.

01:19 So we redraw our diagram and we have the same incident light at an angle of 30 degrees and what we want to do next is we want to find the angle of it refraction.

01:32 So we want to find this angle theta 2.

01:35 So n1, sine of theta 1 is equal to n2, sign of theta 2.

01:42 So n1 is 1 because it's air, and sign of 30 degrees is equal to 1 .523...

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