# A Solution to a Problem That Arises When Studying Refraction and Total Internal Reflection in a Rectangular Glass Slab Using Two Pairs of Pins

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## Introduction: A Solution to a Problem That Arises When Studying Refraction and Total Internal Reflection in a Rectangular Glass Slab Using Two Pairs of Pins

When studying refraction, students usually look at the path that a ray of light takes when it enters (incident ray) and leaves (emergent ray) a rectangular glass slab. Often they perform a simple experiment which involves:

• placing two pins in such a way that the line joining them makes an angle (not a right angle) with one side of a glass slab; the line joining these two pins represents the incident ray;
• placing another pair of pins on the opposite face of the glass slab so that these latter two pins are in line with the first two pins when viewed through the opposite face of the slab; the line joining the second pair of pins represents the emergent ray;
• drawing a line between the following two points:
• the point of intersection of the incident ray and the surface of the glass slab through which the incident ray entered the slab;
• the point of intersection of the emergent ray and the surface of the glass through which the emergent ray left the slab;

this line represents the refracted ray.

This Instructables suggests two other experiments that could be performed with the same equipment so as to study total internal reflection (TIR). The suggestion is based on the concepts of the laws of reflection, refraction (Snell’s law), the critical angle and TIR. For readers not familiar with these topics, refer to these articles and these articles in The Physics Classroom. These topics are used to provide an answer to a problem that arises in such an experiment, namely:

• what is the path taken by a ray of light through a rectangular glass slab when TIR occurs as well as refraction when the only pieces of information available are lines showing:
• the outline of the glass slab;
• the incident ray;
• the emergent ray?

Two possible experiments involving TIR are described in this Instructables:

1. after a ray of light enters the glass slab it undergoes TIR (referred to below as one TIR);
2. after a ray of light enters the glass slab it undergoes TIR followed by another TIR (referred to below as two TIRs).

The first of these experiments involving one TIR has been mentioned in an article without detailing how to find the path taken by a light ray inside the glass slab.

The diagrams shown in the various steps below are based on calculations for which the refractive index of a rectangular glass slab is approximately 1.5, with angles of incidence and refraction approximately equal to 49 degrees and 30 degrees, respectively; these values approximately satisfy the definition of the refractive index of a material, namely:

• refractive index = sin (incident angle)/sin (refractive angle).

## Step 1: The Problem Involving One TIR Occurring Inside the Rectangular Glass Slab.

The solid lines in the above figure show what a result could look like for an experiment illustrating TIR when the pins and glass slab are removed from the surface (e.g. paper) on which the experiment was carried out. The only pieces of information available are straight lines showing:

• an outline of the glass slab (PQRS);
• the incident ray (AB);
• the emergent ray (CD).

As noted in the Introduction, in the classical experiment usually performed by students, the path of the light ray inside the glass slab (the refracted ray) is obtained by drawing a line between the points of intersection of the lines representing the incident and emergent rays with the surfaces of the glass slab (points B and C, respectively). However, this line is not the actual path of a light ray inside the glass slab if TIR takes place.

The problem posed in this Instructables is how to find the path of the light ray inside the glass slab using information based on the physical principles of reflection, refraction and TIR that is usually taught to students studying a first course in optics.

## Step 2: Information Based on the Theory of Reflection, Refraction and TIR.

For an experiment that involves a single TIR occurring in a rectangular glass slab all that is know from theoretical principles is that:

1. the light ray is refracted as it leaves the air surrounding the glass slab and enters the glass slab;
2. the refracted ray then hits the surface of the glass slab that is at right angles to the surface through which the incident ray entered (unless the slab is very wide);
3. since the angle between the refracted ray and the normal to the point of intersection of the refracted ray and the glass surface that is at right angles to the surface through which the incident ray entered the glass slab is greater than the critical angle for glass, the refracted ray then undergoes reflection. In other words, with reference to the glass surface that is at right angles to the surface through which the incident ray entered, the refracted ray is now the incident ray for this surface and according to the laws of reflection the angles between the normal and the incident and reflected rays are equal;
4. the reflected ray reaches the surface of the glass slab opposite the surface through which the incident ray entered and it is refracted as the ray leaves the glass slab and enters the air surrounding the glass slab.

## Step 3: A Solution to the Problem Involving One TIR.

The above diagram shows the expected path taken by the light ray inside the glass slab. BI is the refracted ray after the incident ray enters the glass slab. If the angle made by the refracted ray and its normal BF as it enters the glass slab is r, then. using the information in Step 2 and geometric principles:

1. angle FBI = r;
2. angle FBI = angle BIQ = r (interior angles on opposite sides of the transverse line BI between the parallel lines PQ and IJ are equal);
3. angle BIQ = angle CIR = r (because angle BIJ = angle CIJ by the laws of reflection);
4. angle CIR = angle HCI = r (interior angles on opposite sides of the transverse line IC between the parallel lines IJ and RS are equal).

By Snell’s law that requires the refractive index to be equal to the ratio of the sine of the angle of incidence to the sine of the angle of refraction, it follows that the angle the emergent ray makes with its normal is equal to the angle that the incident ray makes with its normal. Thus, the angle of incidence equals the angle of emergence.

If we let l be the length of the rectangular glass slab (l equals the distance QR), then:

• l = QI + IR.

Since triangles BIQ and CIR are similar,

• BQ/QI = CR/IR.

From which it follows that:

• l = [(BQ/CR) + 1] IR = [(CR/BQ) + 1] QI;

and thus:

• QI = l/[(CR/BQ) + 1];
• IR = l/[(BQ/CR) + 1].

Thus, measuring the distances l, BQ and CR allows one to work out the point at which I should be placed on the side QR of the glass slab and therefore the path of the refracted ray (BI) and the reflected ray (IC) inside the glass slab.

Using the trigonometric relationships for tan r

• tan r = BQ/QI = CR/IR,

it follows that

• l = BQ/tan r + CR/tan r,

so that tan r = (BQ + CR)/l.

Thus, knowing the distances l, BQ and CR allows the angle of refraction of the glass in the rectangular glass slab to be calculated without having to use a protractor to measure angles. Knowing the position of the point I on the side QI of the glass slab also allows sin r to be found from the ratio of the distances BQ over BI. The sine of the angle of incidence i can readily be found without using a protractor using procedures similar to those described in an earlierInstructables. Thus the refractive index of the glass slab follows from the ratio of sin i over sin r.

## Step 4: The Problem and Information Available for Two TIRs Occurring in a Rectanglular Glass Slab.

As in Step 1 the solid lines in the above figure show another type of result that could be obtained for an experiment illustrating TIR when the pins and glass slab are removed and the outline of the glass slab (PQRS), the incident ray (AB) and emergent ray (CD) are drawn.

While the incident and emergent rays turn out to be parallel (shown in the next step) as is found in the classical experiment usually performed by students, if the approach used in the classical experiment of letting the line between points B and C represent the apparent path of the 'refracted' ray of light inside the glass slab, an incorrect result would be obtained for the refractive index of glass (a very large value would be obtained for the refractive index).

In the experiment which involves two TIRs, the first three pieces of information are identical with those listed in Step 2. These three piecesf information are followed by another three pieces of information:

1. the reflected ray then hits the surface of the glass slab opposite that which resulted in the first reflection;
2. since the angle between reflected ray and the normal to the point of intersection of the reflected ray and the glass surface opposite the surface from which the first reflection emanated is greater than the critical angle for glass, the reflected ray is then reflected off the surface that was hit in the previous step according to the laws of reflection;
3. the second reflected ray reaches the surface opposite that through which the incident ray entered the glass slab and is refracted as the ray leaves the glass slab and enters the air surrounding the glass slab.

As before, the only measurements that can be made from these data are the dimensions of the glass slab PQRS (length l and width w) and the distances from the corners of the glass slab to the point where the incident ray entered the glass slab (PB and BQ) and to the point where the emergent ray exited the glass slab (SC and CR); (note that w = PB + BQ = SC + CR). No information at this stage shows the path of the ray of light inside the glass slab.

## Step 5: A Solution to the Problem Involving Two TIRs.

The above diagram shows the expected path taken by the light ray inside the glass slab.

The analysis of this situation follows along the same lines as that performed in Step 3. BI is the refracted ray after the incident ray enters the glass slab. If the angle made by the refracted ray and its normal BF as it enters the glass slab is r, then:

• angle FBI = r;
• angle FBI = angle BIQ = r (interior angles on opposite sides of the transverse line BI between the parallel lines PQ and IJ are equal);
• angle BIQ = angle KIL = r (because angle BIJ = angle KIJ by the laws of reflection);
• angle KIL = angle IKJ = r (interior angles on opposite sides of the transverse line IK between the parallel lines IJ and KL are equal);
• angle IKJ = angle CKS = r (because angle IKL = angle CKL by the laws of reflection);
• angle CKS = angle KCH = r (interior angles on opposite sides of the transverse line CK between the parallel lines KL and RS are equal).

By Snell’s law that requires the refractive index to be equal to the ratio of the sine of the angle of incidence to the sine of the angle of refraction, it follows that the angle the emergent ray makes with its normal is equal to the angle that the incident ray makes with its normal and thus the incident and emergent rays are parallel.

If we let l be the distance QR = PS, then :

• l = QI + IL + KS.

Using the definition of the tangent of an angle and noting that w = KL, we can write:

• tan r = BQ/QI = w/IL = CS/KS.

Therefore we can write l as:

• l = (BQ + w + CS)/tan r,

and express tan r as:

• tan r = (BQ + w + CS)/l.

Thus;

• QI = BQ/tan r = BQ l/(BQ + w + CS)
• IL = w/tan r = w l/(BQ + w + CS)
• KS = CS/tan r = CS l/(BQ + w + CS)

As all the distances on the right hand sides of the above three equations can be measured, one can work out where the point I should be placed on the side QR of the glass slab and where the point K should be placed on the side PS of the glass slab and therefore the path of the refracted ray and the reflected rays inside the glass slab.