(a) Prove that a ray of light incident on the surface of a...

(a) Prove that a ray of light incident on the surface of a sheet of plate glass of thickness $t$ emerges from the opposite face parallel to its initial direction but displaced sideways, as in Fig. $33-69 .$ (b) Show that, for small angles of incidence $\theta,$ this displacement is given by $$x=t \theta \frac{n-1}{n}$$ where $n$ is the index of refraction of the glass and $\theta$ is measured in radians.

(a) Prove that a ray of light incident on the surface of a sheet of plate glass of thickness $t$ emerges from the opposite face parallel to its initial direction but displaced sideways, as in Fig. $33-69 .$
(b) Show that, for small angles of incidence $\theta,$ this displacement is given by
$$x=t \theta \frac{n-1}{n}$$
where $n$ is the index of refraction of the glass and $\theta$ is measured in radians.

Key Concepts

Small-Angle Approximation

The small-angle approximation is a mathematical simplification used when dealing with very small angles, where trigonometric functions such as sine and tangent can be approximated by the angle itself (in radians). This approach streamlines the derivation of formulas, such as the one describing lateral displacement in optical systems, by reducing the complexity of trigonometric expressions.

Index of Refraction

The index of refraction is a measure that quantifies how much light slows down when it enters a medium compared to its speed in a vacuum. It is a central parameter in optics that affects how light bends at interfaces, and it plays a critical role in determining the magnitude of phenomena like the lateral displacement of a light ray in a glass slab.

Lateral Displacement

Lateral displacement refers to the phenomenon where a light ray, after passing through a medium with parallel boundaries, emerges parallel to its original direction yet shifted sideways. This effect is explained by analyzing the geometry of the ray's path through the medium, taking into account the bending at the entry and exit surfaces.

Snell's Law

Snell's Law governs the relationship between the angles of incidence and refraction when light passes from one medium into another with a different refractive index. It states that the product of the refractive index and the sine of the angle of incidence is constant across the interface, which is fundamental in predicting the change in the direction of light rays at boundaries.

Transcript

00:01 In this question, we have a ray of light incident on the last lap.

00:09 So light is shining at this angle.

00:18 Data from the normal and then it gets up and then it comes up and then it comes up like this.

00:27 Okay, so we want to show that in part a, that the light images is from the the opposite phase parallel to its initial direction, now displace sideways, okay? and then show that the x, the displacement of the light ray, is given to be some expression, okay, when the angle is small.

00:59 So to do part a, we'll be using snals law.

01:12 So we have n1, sine data equals to, and 2, sine data 2.

01:21 Okay, so this is our data 1, and then this is our data 2.

01:28 This is also our data 2, and then, so at the left interface, so this is the left interface, and then this is the right interface.

01:48 We have 1 times sine data goes 2.

01:52 And side data 2 and then at the so we have side data equals to n side data 2 at the end then for the right interface we have n sine data 2 equals to sign data again okay so so let's say we don't know this angle data we call it data 3 right and so side data it goes to side data 3 which means that data 3 is equal to data and this shows that the emergent ray is parallel to the incident ray on the right side on the left side sorry okay and then to manage to show that is parallel to the incident ray and from diagram is is going to be displaced okay so that's for part a then part b.

03:15 We need to show that for small angle of incidence, we can show the displacement to be to be of certain form.

03:25 So i'm just going to exaggerate the figure a little bit.

03:31 Okay.

03:33 So you have this angle coming in.

03:39 Then this is a angle.

03:43 It is a refracted ray and then comes out.

03:47 So if you compare the we draw the incident ray.

03:57 We want to calculate this x.

04:00 So we need to label some angles...