Why a Prism Spreads Colours but a Glass Slab Usually Doesn’t

A narrow beam of white light passes through a triangular glass prism and spreads into a band of colours. Send the same beam through a rectangular glass slab, and the light usually emerges looking almost white. Both objects are made of glass. Both refract light at two surfaces. So why does one produce a visible spectrum while the other usually doesn’t?

The difference is the orientation of their surfaces. A prism has non-parallel refracting faces, so different colours emerge travelling in different directions. A parallel-sided glass slab also refracts different colours by different amounts, but its second surface restores their original direction. The colours can still be displaced sideways by slightly different amounts.

Here’s the useful prediction: move a screen farther from each object. The spectrum from the prism generally gets wider. The colour separation caused by an ideal parallel slab stays the same after the light has emerged.

01The experiment: same light, different glass shape

Imagine setting up a simple investigation with a white-light source, a narrow slit, a collimating lens, a glass sample, and a movable white screen.

The slit produces a narrow beam. The lens makes the rays approximately parallel, reducing the spread caused by light leaving the slit in different directions.

Arrange the equipment in this order:

1. A white-light source illuminates a narrow slit.
2. A collimating lens produces an approximately parallel beam.
3. The beam passes through either a triangular prism or a parallel-sided glass slab.
4. A white screen intercepts the transmitted light.
5. The screen is moved farther away to investigate how the coloured pattern changes.

What should happen?

Suppose the beam contains red and blue light. You might predict that both pieces of glass should separate them, because blue light generally refracts more strongly than red light.

That prediction is partly correct.

At the first air-to-glass boundary, both shapes separate the directions of the red and blue rays. But what happens at the second boundary decides whether that angular separation survives.

These are predictions from an idealised ray model, not measurements from a reported investigation. Actual results depend on beam width, glass dispersion, geometry, and screen position.

Why improve the original setup?

A coloured patch on one screen doesn’t tell us exactly what caused the separation.

A broad beam could already contain rays travelling in slightly different directions. An irregular glass surface could redirect them. A single screen also cannot distinguish a sideways displacement from a difference in outgoing direction.

The improved test uses a narrow, approximately collimated beam and measures the coloured positions at two screen distances.

If the red-blue separation grows with distance, the emerging rays have different directions. If it remains constant, they are parallel but may occupy different positions.

That distinction is the key to the prism-versus-slab problem.

02What Newton’s prism experiment established

Isaac Newton faced a more fundamental question in the 1660s: did a prism create colours, or did white light already contain components that refracted differently?

In an account published in 1672, Newton described allowing sunlight through a small hole in a darkened window. A triangular glass prism intercepted the beam and refracted it towards the opposite wall.

He expected a roughly circular patch based on the ordinary geometry of refraction. Instead, he observed an elongated, coloured image.

Why was it so long?

The first observation couldn’t settle the question. Perhaps different parts of the incoming sunlight struck the prism at different angles. Perhaps the glass was imperfect. Newton tested several possible explanations, including changing the aperture and using a second prism oriented to oppose the first.

He found that an oppositely oriented second prism could restore the elongated image to an approximately circular one. This challenged the idea that random glass imperfections caused the length.

The crucial revision: selecting one part of the spectrum

Newton then described a more discriminating arrangement, which he called his experimentum crucis.

The light travelled through the first prism and a hole in a nearby board, then across the room towards another board with a small opening. A second prism sat beyond that opening, with a wall farther away.

By slowly rotating the first prism, Newton could make different portions of its spectrum pass through the second opening.

He observed that portions selected from different ends of the spectrum were refracted by different amounts by the second prism.

This supported the conclusion that the original white light contained components with different refrangibilities: they did not all respond identically to the same refraction.

The revised apparatus helped distinguish different refrangibility from the simpler explanation that the long image was caused only by different incident directions.

Newton did not need to compare a rectangular slab with a prism to establish this result. That comparison is a useful modern teaching experiment, while the historical evidence concerned the composition of white light.

His description is available in Newton’s published account of light and colours.

Modern physics explains the result using wavelength-dependent refractive index.

03Why glass refracts different colours differently

Think of red and blue rays entering glass from air at the same angle.

Both slow down, but their speeds in the glass are slightly different. In ordinary transparent glass across the visible spectrum, blue light generally has a higher refractive index than red light.

The refractive index is

\[ n=\frac{c}{v} \]

where:

• \(n\) is refractive index, with no units.
• \(c\) is the speed of light in a vacuum, in \(\mathrm{m\,s^{-1}}\).
• \(v\) is the phase speed of light in the material, in \(\mathrm{m\,s^{-1}}\).

The dependence of refractive index on wavelength is called dispersion.

For ordinary glass in the visible range, a shorter wavelength generally means a larger refractive index.

Draw the normal before following the ray

A normal is an imaginary line perpendicular to a surface at the point where light meets it.

Every incidence or refraction angle is measured from that local normal, not from the glass surface.

For example, suppose an incident ray makes \(35^\circ\) with the glass surface.

Its angle of incidence is \(55^\circ\), not \(35^\circ\), because the normal is perpendicular to the surface.

The mathematical rule is Snell’s law:

\[ n_1\sin i=n_2\sin r \]

Here, \(n_1\) is the refractive index of the incident medium, \(n_2\) is that of the transmitted medium, \(i\) is the incident angle, and \(r\) is the refracted angle.

For light entering glass from air, a larger glass refractive index produces a smaller refracted angle.

This means blue bends closer to the normal than red.

For the calculations below, we’ll use these illustrative values.

These are assumed values for teaching, not measured data for a particular glass. We assume homogeneous, transparent, isotropic materials, flat surfaces, and negligible absorption, scattering, and diffraction. Unless stated otherwise, the beam is collimated and air surrounds the glass.

Local ray angles are taken as positive angles between \(0^\circ\) and \(90^\circ\), measured from the relevant normal. Angular separations are reported as positive magnitudes.

04Why the second surface changes everything

Parallel-sided slab: the outgoing angles become equal again

Imagine a ray travelling from left to right through a rectangular glass block. Its entry and exit faces are parallel.

At the entry surface, blue bends closer to the normal than red. The rays follow slightly different straight paths through the glass.

At the second surface, both colours travel from glass back into air. They bend away from the normal.

Here’s what makes the slab special: its two surface normals are parallel.

Let \(i\) be the original incident angle, \(r_\lambda\) the angle inside the glass for a particular wavelength, and \(e_\lambda\) the emerging angle.

At the first surface:

\[ n_{\mathrm{air}}\sin i =n_{\mathrm{glass},\lambda}\sin r_\lambda \]

At the second surface, the internal angle of incidence is still \(r_\lambda\):

\[ n_{\mathrm{glass},\lambda}\sin r_\lambda =n_{\mathrm{air}}\sin e_\lambda \]

Combining them gives

\[ \sin e_\lambda=\sin i \]

Since both angles lie between \(0^\circ\) and \(90^\circ\),

\[ e_\lambda=i \]

This is true for every transmitted wavelength.

The red ray emerges parallel to its original direction. So does the blue ray.

However, they haven’t followed the same path inside the glass. Their exit positions differ slightly.

The second refraction cancels the angular difference, not necessarily the sideways separation.

If the beam enters perpendicular to the faces, neither colour bends at either surface, and there is no geometrical lateral displacement.

For more on tracing individual paths, see How to Trace a Ray Through a Parallel-Sided Glass Block.

05Prism: the second surface does not undo the angular spread

Now replace the slab with a triangular glass prism.

The first surface behaves as before: blue bends closer to its normal than red.

But the second surface is tilted relative to the first. Its normal points in a different direction.

The red and blue rays therefore meet the second face at different angles, and their outgoing directions are not restored to the original incident direction.

For a normally dispersive glass, blue experiences greater overall deviation towards the prism’s base than red.

For the usual transmitted-ray geometry, define:

• \(A\): angle between the two refracting faces, called the prism angle.
• \(i\): incidence angle at the first face.
• \(r_1\): refracted angle inside the first face.
• \(r_2\): internal incidence angle at the second face.
• \(e\): emergence angle from the second face.
• \(\delta\): total angular deviation from the original beam direction towards the base.

The prism’s geometry gives

\[ r_1+r_2=A \]

and its total deviation is

\[ \delta=i+e-A \]

These relationships apply to the ordinary prism path crossing both refracting faces. A transmitted ray must also satisfy Snell’s law at each boundary.

The important difference from a slab is that \(r_2\) is not generally equal to \(r_1\).

Because the normals are not parallel, the two refractions do not cancel the change in direction.

06Does a parallel glass slab ever show colours?

Yes. There are three important qualifications.

1. A sufficiently thick slab can separate narrow beams

The wavelength-dependent lateral displacement increases with slab thickness.

For the same incidence angle and refractive indices used earlier,

\[ \Delta y=t(\tan r_R-\tan r_B) \]

Doubling \(t\) doubles the separation.

If a slab is thick enough, and the incident beam is sufficiently narrow and well collimated, different wavelengths can emerge at measurably different positions. They remain parallel once they leave.

The lack of a wide spectrum in an ordinary demonstration is therefore a practical outcome of small separation and overlapping light, not proof that no dispersion occurs.

2. A prism does not always produce an obvious rainbow

A prism illuminated by a broad, divergent white beam may produce overlapping coloured regions rather than distinct bands.

The screen also needs enough distance for the angular separation to become visible.

At normal incidence on a prism’s first face, no colour is bent at entry, but the second face can still separate the colours. This is because its normal is tilted.

3. The surrounding media and internal boundaries matter

The exact parallel-ray result assumes a uniform slab, parallel boundaries, and the same surrounding medium on both sides.

If light enters from air but exits into a different medium with a wavelength-dependent refractive index, the outgoing directions need not match the original direction.

Similarly, a transparent device with parallel outer faces can still produce angular dispersion if it contains a tilted internal boundary between different materials.

So the reliable rule is not simply ‘parallel outside faces mean no spectrum’. You must follow every boundary and identify the media on each side.

A useful diagnostic

The next questions test how far these conclusions can be pushed.

07Questions and solutions

Question 1

A narrow white beam enters a parallel-sided glass block from air at \(35^\circ\) to the entry normal.

Assume the refractive index is 1.50 for red and 1.52 for blue.

a. Calculate the internal refraction angle for each colour.

b. Find the emergence angle for each colour when the beam returns to air.

c. Which colour is displaced farther from the straight-through path, assuming the slab has a non-zero thickness?

Solution 1

Both colours emerge at \(35^\circ\) to the exit normal, but blue is displaced farther sideways than red.

For part (a), use Snell’s law:

\[ n_{\mathrm{air}}\sin i=n_{\mathrm{glass}}\sin r \]

For red,

\[ \begin{aligned} r_R&=\sin^{-1}\left(\frac{\sin35^\circ}{1.50}\right)\\ &=22.48^\circ \end{aligned} \]

For blue,

\[ \begin{aligned} r_B&=\sin^{-1}\left(\frac{\sin35^\circ}{1.52}\right)\\ &=22.17^\circ \end{aligned} \]

For part (b), both colours emerge back into air through a surface parallel to the first.

Therefore,

\[ e_R=e_B=35^\circ \]

For part (c), blue travels closer to the normal inside the slab, so it covers less distance parallel to the faces before exiting.

It consequently has a larger lateral displacement from the straight-through path.

The tempting mistake is to predict a larger emergence angle for blue simply because it refracts more strongly. That ignores the geometry of the second boundary.

Question 2

A collimated beam contains overlapping red and blue components, each with a width of \(2.0\ \mathrm{mm}\), measured parallel to the face of a glass slab.

The beam enters a parallel-sided slab at \(45^\circ\). The slab is \(150\ \mathrm{mm}\) thick, measured perpendicular to its faces.

Use \(n_R=1.50\) and \(n_B=1.52\), with air on both sides.

Assume uniform beam widths, sharp edges, and negligible diffraction or absorption.

a. Calculate the separation between the red and blue beam centres at the exit face.

b. Are their illuminated bands completely separate?

c. Would moving a screen from \(0.5\ \mathrm{m}\) to \(5.0\ \mathrm{m}\) beyond the slab make the bands completely separate?

d. Calculate the minimum slab thickness needed for the red and blue bands to become just disjoint in this idealised model.

Solution 2

The red and blue beam centres emerge approximately \(1.35\ \mathrm{mm}\) apart. Their \(2.0\ \mathrm{mm}\)-wide bands still overlap, and moving the screen cannot remove that overlap. A slab thickness of about \(223\ \mathrm{mm}\) is needed for the bands to become disjoint.

For part (a), the internal refraction angles are

\[ r_R=28.13^\circ,\qquad r_B=27.72^\circ \]

The separation is

\[ \Delta y=t(\tan r_R-\tan r_B) \]

Substitute \(t=150\ \mathrm{mm}\):

\[ \begin{aligned} \Delta y &=150(\tan28.13^\circ\\ &\quad-\tan27.72^\circ)\\ &\approx1.35\ \mathrm{mm} \end{aligned} \]

For part (b), the illuminated bands each occupy \(2.0\ \mathrm{mm}\) along the relevant transverse direction.

Their overlap is

\[ 2.0-1.35=0.65\ \mathrm{mm} \]

They are not fully separated.

For part (c), the slab restores both colours to the same propagation direction. The difference in their positions remains \(1.35\ \mathrm{mm}\) after emergence.

Moving the screen cannot increase this separation.

For part (d), the bands become just disjoint when their centre separation equals their width:

\[ \Delta y=2.0\ \mathrm{mm} \]

So

\[ t(\tan28.13^\circ-\tan27.72^\circ)=2.0 \]

Rearranging and using unrounded angle values,

\[ t\approx222.4\ \mathrm{mm} \]

A thickness of about \(223\ \mathrm{mm}\) is sufficient.

The tempting assumption is that any dispersive glass will eventually show a wider spectrum if the screen is far enough away. That works for angular dispersion, but not for two parallel emerging beams. Here, increasing glass thickness is what increases the colour separation.

Question 3

An unknown transparent optical component receives a narrow, collimated red-blue beam in air.

Two screens, each perpendicular to the mean emerging beam direction, are used to measure the separation between the red and blue spot centres.

The following data are illustrative, not experimental measurements.

The blue spot remains on the same side of the red spot at both positions. Assume straight propagation in air, negligible beam divergence, and small angular separation.

a. Estimate the angular separation between the emerging rays.

b. Estimate the separation at the exit plane.

c. Predict the separation at \(4.00\ \mathrm{m}\).

d. The technician has two candidate components: a uniform parallel-sided glass slab surrounded by air, or a triangular glass prism. Which is consistent with the data? Can the measurements establish the component’s exact shape?

Solution 3

The rays differ in direction by approximately \(0.29^\circ\), and their positions were already about \(0.50\ \mathrm{mm}\) apart at the exit plane. Of the two candidate components, a prism is consistent with the data; an ideal uniform parallel-sided slab in air is not.

For part (a), the separation increases by

\[ 13.0-3.0=10.0\ \mathrm{mm} \]

over a screen displacement of

\[ 2.50-0.50=2.00\ \mathrm{m} \]

The rate of separation is

\[ \frac{0.0100}{2.00} =0.00500\ \mathrm{m/m} \]

Using the small-angle approximation,

\[ \begin{aligned} \Delta\theta &\approx\tan^{-1}(0.00500)\\ &=0.286^\circ \end{aligned} \]

For part (b), let \(\Delta y_0\) be the signed exit-plane separation. Using metres for distance and millimetres for separation,

\[ 3.0=\Delta y_0+5.0(0.50) \]

Therefore,

\[ \Delta y_0=0.50\ \mathrm{mm} \]

For part (c),

\[ \begin{aligned} \Delta y &=0.50+5.0(4.00)\\ &=20.5\ \mathrm{mm} \end{aligned} \]

For part (d), the growing separation demonstrates that the emerging rays are not parallel.

A single uniform slab with parallel faces and air on both sides cannot produce that angular difference. A prism can.

However, the measurements alone do not prove the component is a prism. Another device containing tilted refracting boundaries could produce a similar result.

The trap is treating the separation at one screen as purely angular dispersion. The actual separation contains both an initial positional offset and a contribution from different outgoing directions.

Question 4

A glass prism has a refracting angle of \(41.5^\circ\).

A narrow beam containing red and blue light strikes its first face at normal incidence. The prism is surrounded by air.

Assume:

• Red-light refractive index: \(1.50\)
• Blue-light refractive index: \(1.52\)
• Air refractive index: \(1.00\)

The beam travels directly towards the second refracting face.

a. Find the angle of incidence at the second face for both colours.

b. Determine whether each colour can emerge through that face.

c. Calculate the emergence angle for any transmitted colour.

d. Explain why the result challenges the usual statement that a prism simply spreads blue farther than red.

Solution 4

Red light emerges through the second face at approximately \(83.7^\circ\) to its normal, while blue light undergoes total internal reflection at that face.

For part (a), normal incidence at the first face means

\[ r_1=0^\circ \]

Using the prism geometry,

\[ \begin{aligned} r_2&=A-r_1\\ &=41.5^\circ \end{aligned} \]

Both colours reach the second surface at \(41.5^\circ\).

For part (b), light travels from glass into air, so total internal reflection is possible.

The critical angle is

\[ \theta_c=\sin^{-1}\left(\frac{1.00}{n}\right) \]

For red,

\[ \theta_{cR}=41.81^\circ \]

For blue,

\[ \theta_{cB}=41.14^\circ \]

Compare each with the actual incidence angle:

• Red: \(41.5^\circ<41.81^\circ\), so refraction is possible.
• Blue: \(41.5^\circ>41.14^\circ\), so total internal reflection occurs.

For part (c), apply Snell’s law to red:

\[ \sin e_R=1.50\sin41.5^\circ \]

Hence,

\[ e_R\approx83.7^\circ \]

The red ray emerges almost parallel to the surface.

For blue, Snell’s law would require

\[ \sin e_B=1.52\sin41.5^\circ \]

This gives a value greater than 1, so there is no real transmitted ray angle. Blue reflects internally instead.

For part (d), greater refraction does not guarantee that a colour will emerge.

The second boundary places a physical limit on transmission. A more strongly refracted colour may reach the condition for total internal reflection first.

Blue might subsequently escape through another face after further reflections, but it does not transmit through this second face on the encounter described.

The mistaken approach is to use Snell’s law to calculate an emergence angle without first checking whether one exists.

Question 5

An optical engineer constructs a transparent cell with perfectly parallel outer faces.

The entry side contains one type of glass. The exit side contains a transparent liquid. The two materials meet at a single flat internal boundary whose normal is tilted \(25.0^\circ\) above the common outward direction of the outer-face normals.

A narrow red-blue beam enters from air perpendicular to the first outer face.

The refractive indices are:

Assume all interfaces are flat, the ray reaches each intended boundary, and all materials are transparent and homogeneous. Ignore reflections, diffraction, and absorption.

Two students disagree.

Student A argues that the red and blue rays must emerge parallel to the original beam because the outer faces are parallel and air surrounds both sides.

Student B argues that the angled internal boundary can cause angular colour separation, even though the outer faces are parallel.

a. Which student is correct?

b. Calculate the final propagation direction of each colour relative to the original beam, giving the angular separation to two significant figures.

c. Would placing the screen farther away increase the separation of the red and blue spots?

d. Give one change to the internal optical arrangement that would eliminate this angular colour separation, without changing the orientation of the outer faces.

Solution 5

Student B is correct. Red light emerges along the original direction, but blue light emerges approximately \(0.54^\circ\) below it. Moving the screen farther away increases the spatial separation.

The decisive feature is the tilted internal glass-to-liquid boundary, not the parallel outer faces.

For part (a), the incoming ray strikes the first outer surface at normal incidence.

Neither colour changes direction there, even though the refractive indices differ.

Both rays then meet the internal boundary at \(25.0^\circ\) to its normal.

Now the medium pair matters.

For red light,

\[ n_{\mathrm{glass},R} =n_{\mathrm{liquid},R} =1.50 \]

Its refractive index does not change at the internal boundary, so its direction remains unchanged.

Blue light has a different situation:

\[ n_{\mathrm{glass},B}=1.52 \]

and

\[ n_{\mathrm{liquid},B}=1.50 \]

It crosses from a higher-index material into a lower-index material at an oblique angle. It must bend away from the internal normal.

This immediately shows why Student A’s cancellation argument fails. The angled internal interface has changed blue light’s direction before it reaches the final outer face.

For part (b), apply Snell’s law at the internal interface for blue:

\[ 1.52\sin25.0^\circ =1.50\sin r_B \]

Therefore,

\[ \begin{aligned} r_B &=\sin^{-1}\left( \frac{1.52\sin25.0^\circ}{1.50} \right)\\ &=25.357^\circ \end{aligned} \]

The internal normal was tilted \(25.0^\circ\) above the original ray. Blue bends away from that normal, so its direction in the liquid is below the original direction by

\[ \begin{aligned} \phi&=25.357^\circ-25.0^\circ\\ &=0.357^\circ \end{aligned} \]

Now consider the final liquid-to-air boundary.

The outer face is perpendicular to the original beam, so blue reaches it at \(0.357^\circ\) to its normal.

Using Snell’s law again,

\[ 1.50\sin0.357^\circ =1.00\sin e_B \]

Thus,

\[ e_B\approx0.535^\circ \]

Red meets the final surface at normal incidence and continues undeviated:

\[ e_R=0^\circ \]

So the outgoing rays differ by approximately

\[ \Delta e=0.54^\circ \]

Blue travels below the original beam direction, while red continues straight ahead.

For part (c), the two rays now have different directions in the same surrounding air.

Their positional separation therefore changes as they propagate. At sufficiently large distances, the angular contribution dominates, so moving the screen farther away increases their separation.

For part (d), one solution is to replace the liquid with a transparent material whose refractive index matches the glass at both wavelengths:

\[ n_{\mathrm{liquid},R}=1.50 \]

\[ n_{\mathrm{liquid},B}=1.52 \]

Neither colour would then refract at the internal interface. Both would reach the final outer face at normal incidence and emerge along the original beam direction.

Another solution is to make the internal boundary parallel to the outer faces. With the same medium on entry and exit, all relevant interface normals would then be parallel, restoring the angular cancellation.

The strongest trap is applying the parallel-slab rule to a device merely because its outside looks like a slab.

The rule depends on the complete sequence of refracting boundaries, their orientations, and the refractive indices of the materials.

Here, parallel outer faces are not sufficient. The angled internal boundary creates a wavelength-dependent change in direction that the final surface does not cancel.

08What to remember when comparing a prism and a slab

A prism and a parallel slab both produce wavelength-dependent refraction.

Their different outcomes follow from the geometry:

• A uniform parallel-sided slab surrounded by air restores the original propagation direction for each transmitted colour, while allowing small wavelength-dependent sideways displacements.
• A prism normally produces different emerging directions for different colours, so a distant screen can reveal a broad spectrum.
• The visibility of colours also depends on beam width, screen distance, material dispersion, and whether the light can transmit through every boundary.

When solving an unfamiliar refraction problem, identify the medium pair at each surface, draw its normal, measure the local incidence angle, and apply Snell’s law. Don’t assume that reaching a glass-to-air boundary guarantees transmission.

That last decision leads directly to How to Decide if a Ray Refracts or Reflects at a Second Boundary, especially for prisms whose geometry brings a colour close to its critical angle.