How to Calculate Wavelength Inside Glass from Vacuum Wavelength

A monochromatic light source has a vacuum wavelength of 600 nm. It enters a glass block with a refractive index of 1.50. Does the light keep its 600 nm wavelength inside the glass?

No. Its wavelength decreases to 400 nm.

The calculation is straightforward:

\[ \lambda_{\text{glass}}=\frac{\lambda_0}{n} =\frac{600}{1.50}=400\text{ nm} \]

The key is to divide the vacuum wavelength by the refractive index of the glass. The light’s frequency stays constant as it crosses the boundary, but its speed and wavelength change.

01The formula for wavelength inside glass

The relationship you need is:

\[ \boxed{\lambda_{\text{glass}}=\frac{\lambda_0}{n_{\text{glass}}}} \]

Where:

• \(\lambda_{\text{glass}}\) is the wavelength inside the glass, measured in metres or nanometres.
• \(\lambda_0\) is the vacuum wavelength of the light.
• \(n_{\text{glass}}\) is the absolute refractive index of the glass, measured relative to vacuum.

The refractive index is defined as:

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

Here, \(c\) is the speed of light in a vacuum, approximately \(3.00\times10^8\text{ m s}^{-1}\), and \(v\) is the speed of light in the material.

Why does this give us the wavelength?

The wave equation is:

\[ v=f\lambda \]

Rearranging:

\[ \lambda=\frac{v}{f} \]

At a stationary boundary between vacuum and glass, the frequency \(f\) of the transmitted light remains unchanged.

Inside glass, the speed becomes \(c/n\). Therefore:

\[ \begin{aligned} \lambda_{\text{glass}}&=\frac{c/n}{f}\\ &=\frac{c}{f}\times\frac{1}{n}\\ &=\frac{\lambda_0}{n} \end{aligned} \]

That’s where the formula comes from.

Why does the wavelength decrease?

Think about what happens to the wave as it enters glass.

1. The light travels more slowly in glass than in a vacuum.
2. Its frequency stays constant.
3. The distance between successive wave crests decreases.

Since wavelength is the distance between successive crests, a lower wave speed at the same frequency means a shorter wavelength.

An important distinction: the light does not change colour simply because its wavelength becomes shorter inside the glass. Its frequency determines its colour, and that frequency remains unchanged.

02How an experiment helps you find the refractive index

What if a question doesn’t give you the refractive index of the glass?

You may need to determine it from measurements of refraction.

Imagine directing a narrow laser beam from air into a transparent glass block. The beam changes direction at the boundary because it travels at a different speed in the glass.

The experiment uses:

• A narrow beam from a ray box or low-power laser.
• A transparent glass block.
• A sheet of paper.
• A ruler and protractor.

The beam travels from the source through air, meets the first surface of the glass, and continues through the material. You trace its path and measure the angles at the first boundary.

Draw the normal before measuring angles

The normal is an imaginary line drawn perpendicular to the boundary at the point where the light ray strikes it.

There are two angles to identify:

• Angle of incidence, \(i\): the angle between the incoming ray and the normal.
• Angle of refraction, \(r\): the angle between the refracted ray and the normal.

Both angles are measured from the normal, not from the surface.

For example, if a ray makes an angle of \(60^\circ\) with the glass surface, its angle of incidence is:

\[ i=90^\circ-60^\circ=30^\circ \]

This distinction is essential when using Snell’s law.

Use Snell’s law to find the refractive index

Snell’s law is:

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

Where:

• \(n_1\) is the refractive index of the initial medium.
• \(n_2\) is the refractive index of the medium the light enters.
• \(i\) is the angle of incidence.
• \(r\) is the angle of refraction.

For light travelling from air into glass:

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

Rearranging:

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

Since air has a refractive index very close to 1.00, HSC calculations usually approximate:

\[ n_{\text{glass}}\approx\frac{\sin i}{\sin r} \]

You can then use this refractive index in the wavelength formula.

The experiment predicts that the ray bends towards the normal when entering ordinary glass from air. The measured refraction angle should therefore be smaller than the incidence angle.

One limitation is that a single pair of angle measurements can contain considerable experimental uncertainty. Repeating the measurements at different incidence angles and comparing the results gives a more reliable estimate.

Also, don’t mistake the emergent ray from the opposite face of a rectangular block for the refracted ray inside the glass. For parallel faces, the emergent ray is parallel to the original incident ray, but it is laterally displaced.

For a fuller explanation of collecting multiple measurements and finding refractive index from a sine-sine graph, see How to Find Refractive Index from a Sine-Sine Graph.

03Worked examples

04What happens when light leaves glass?

So far, we’ve considered light entering glass from air. Now reverse the direction.

Suppose light travels through glass with a refractive index of 1.60 and reaches a glass-air boundary.

Its vacuum wavelength is 520 nm.

First, calculate its wavelength inside glass:

\[ \lambda_{\text{glass}}=\frac{520}{1.60} =325\text{ nm} \]

When it leaves the glass and enters air, its speed increases again. Its wavelength therefore increases.

Assuming \(n_{\text{air}}\approx1.00\):

\[ \lambda_{\text{air}}\approx520\text{ nm} \]

The wavelength returns to approximately its vacuum value.

The frequency has not changed at either boundary.

This gives us a useful rule:

• Air to glass: speed decreases and wavelength decreases.
• Glass to air: speed increases and wavelength increases.

The direction of the ray also changes. It bends towards the normal when entering a higher-index material and away from the normal when entering a lower-index material, provided that a transmitted ray exists.

Absolute versus relative refractive index

There is one more detail worth getting right.

The formula:

\[ \lambda_{\text{glass}}=\frac{\lambda_0}{n_{\text{glass}}} \]

uses the absolute refractive index of glass, measured relative to vacuum.

However, some questions give a relative refractive index between two materials.

The relative refractive index of medium 2 compared with medium 1 is:

\[ n_{2/1}=\frac{n_2}{n_1} \]

If the wavelength in medium 1 is already known, you can use:

\[ \boxed{\lambda_2=\frac{\lambda_1}{n_{2/1}}} \]

Equivalently:

\[ \lambda_2=\lambda_1\frac{n_1}{n_2} \]

Keep the reference medium consistent.

For typical HSC calculations, air is approximated as having a refractive index of 1.00. If a question supplies a more precise value, use it.

One final limitation: refractive index depends on the wavelength of the light. Different colours can have slightly different refractive indices in the same glass. Always use the refractive index specified for the relevant wavelength.

05A reliable method for exam questions

When a question asks you to find the wavelength inside glass from a vacuum wavelength, work through these decisions.

1. Identify the wavelength you have been given. Is it a vacuum wavelength or a wavelength in another material?
2. Identify the target medium. In this case, it is glass.
3. Check whether the refractive index is absolute or relative to another material.
4. If you have a vacuum wavelength and an absolute refractive index, divide the wavelength by the refractive index.
5. If you need to calculate the refractive index from angles, draw the normal and apply Snell’s law.
6. Check that your answer makes physical sense. For ordinary glass with \(n>1\), the wavelength should be shorter than the vacuum wavelength.

The most common mistakes are:

• Multiplying the vacuum wavelength by the refractive index instead of dividing.
• Measuring angles from the surface rather than the normal.
• Confusing the angle of emergence with the angle of refraction.
• Assuming that an unchanged frequency means an unchanged wavelength.
• Using a relative refractive index with a wavelength measured in a different reference medium.

The questions below build on these ideas, starting with a direct calculation and progressing to situations involving multiple boundaries.

06Questions and solutions

Question 1

A monochromatic light source has a vacuum wavelength of 510 nm. It enters a transparent glass block with an absolute refractive index of 1.50.

Calculate the wavelength of the light inside the glass.

Solution 1

The wavelength inside the glass is 340 nm.

We are given the vacuum wavelength and the absolute refractive index, so we can directly apply:

\[ \lambda_{\text{glass}}=\frac{\lambda_0}{n_{\text{glass}}} \]

Substitute:

\[ \begin{aligned} \lambda_{\text{glass}} &=\frac{510}{1.50}\\ &=340\text{ nm} \end{aligned} \]

Therefore:

\[ \boxed{\lambda_{\text{glass}}=340\text{ nm}} \]

The wavelength decreases because light travels more slowly in glass. Its frequency remains unchanged.

Question 2

A monochromatic laser has a vacuum wavelength of 600 nm. It enters a glass block from air.

The incident ray makes an angle of \(65^\circ\) with the glass surface. The refracted ray makes an angle of \(16.5^\circ\) with the normal.

Assume the refractive index of air is 1.00.

Calculate:

(a) The angle of incidence.

(b) The refractive index of the glass.

(c) The wavelength of the light inside the glass.

Solution 2

The wavelength inside the glass is approximately 403 nm, and the refractive index is approximately 1.49.

The important first step is converting the given surface angle into an angle measured from the normal.

(a) Find the angle of incidence.

The normal is perpendicular to the glass surface.

Therefore:

\[ \begin{aligned} i&=90^\circ-65^\circ\\ &=25^\circ \end{aligned} \]

(b) Calculate the refractive index.

The light travels from air into glass, so Snell’s law gives:

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

Rearrange:

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

Substitute:

\[ \begin{aligned} n_{\text{glass}} &=\frac{1.00\sin25^\circ}{\sin16.5^\circ}\\ &=\frac{0.4226}{0.2840}\\ &=1.488 \end{aligned} \]

Therefore:

\[ n_{\text{glass}}\approx1.49 \]

(c) Calculate the wavelength inside glass.

Use:

\[ \lambda_{\text{glass}}=\frac{\lambda_0}{n_{\text{glass}}} \]

\[ \begin{aligned} \lambda_{\text{glass}} &=\frac{600}{1.488}\\ &=403.2\text{ nm} \end{aligned} \]

Therefore:

\[ \boxed{\lambda_{\text{glass}}\approx403\text{ nm}} \]

A common mistake would be to substitute \(65^\circ\) directly into Snell’s law. That angle is measured from the surface, not the normal. The correct incidence angle is \(25^\circ\).

Question 3

A monochromatic laser with a vacuum wavelength of 520 nm travels through glass with an absolute refractive index of 1.60.

The ray reaches a glass-air boundary at an angle of incidence of \(30^\circ\) to the normal.

Assume the refractive index of air is 1.00.

Calculate:

(a) The wavelength inside the glass.

(b) The angle of refraction in air.

(c) The wavelength in air after the light leaves the glass.

Explain why the frequency does not change.

Solution 3

The wavelength inside glass is 325 nm. The ray refracts into air at approximately \(53.1^\circ\) to the normal, where its wavelength becomes approximately 520 nm.

(a) Find the wavelength inside glass.

Use the absolute refractive index:

\[ \lambda_{\text{glass}}=\frac{\lambda_0}{n_{\text{glass}}} \]

Substitute:

\[ \begin{aligned} \lambda_{\text{glass}} &=\frac{520}{1.60}\\ &=325\text{ nm} \end{aligned} \]

Therefore:

\[ \boxed{\lambda_{\text{glass}}=325\text{ nm}} \]

(b) Find the angle of refraction in air.

The light travels from glass into air.

Apply Snell’s law:

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

Rearrange:

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

Substitute:

\[ \begin{aligned} \sin r &=\frac{1.60\sin30^\circ}{1.00}\\ &=1.60(0.5)\\ &=0.80 \end{aligned} \]

Take the inverse sine:

\[ r=\sin^{-1}(0.80) \]

\[ \boxed{r=53.1^\circ} \]

The ray bends away from the normal as it enters the lower-index medium.

(c) Find the wavelength in air.

The wavelength in air is approximately equal to the vacuum wavelength:

\[ \boxed{\lambda_{\text{air}}\approx520\text{ nm}} \]

The wavelength increases from 325 nm inside glass to approximately 520 nm in air.

Why does frequency remain unchanged?

The boundary is stationary, so the transmitted wave has the same frequency as the incoming wave.

The speed increases when light enters air, and the wavelength increases accordingly:

\[ v=f\lambda \]

The change in wavelength does not require a change in frequency.

Question 4

An experimental group wants to determine the refractive index of an unknown transparent glass sample.

They use a monochromatic laser with a vacuum wavelength of 589 nm. The laser travels from air into the glass.

The group records the following illustrative experimental data. All angles are measured from the normal.

Assume the refractive index of air is 1.00.

(a) Calculate the refractive index of the glass for each trial.

(b) Determine the average refractive index.

(c) Calculate the wavelength inside the glass using the average refractive index.

(d) Explain why the group should repeat its measurements rather than rely on one trial. Also explain why measuring the emergent ray from the opposite face of a rectangular block would give an incorrect result if it were treated as the internal refracted ray.

Solution 4

The average refractive index is approximately 1.50, giving a wavelength inside the glass of approximately 393 nm.

(a) Calculate the refractive index for each trial.

Use Snell’s law:

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

Rearrange:

\[ n_{\text{glass}}=\frac{\sin i}{\sin r} \]

For Trial 1:

\[ n_1=\frac{\sin30^\circ}{\sin19.5^\circ} \]

\[ n_1\approx1.498 \]

For Trial 2:

\[ n_2=\frac{\sin45^\circ}{\sin28.1^\circ} \]

\[ n_2\approx1.501 \]

For Trial 3:

\[ n_3=\frac{\sin60^\circ}{\sin35.3^\circ} \]

\[ n_3\approx1.499 \]

(b) Calculate the average refractive index.

\[ \begin{aligned} n_{\text{average}} &=\frac{1.498+1.501+1.499}{3}\\ &=1.499 \end{aligned} \]

Therefore:

\[ \boxed{n_{\text{glass}}\approx1.50} \]

(c) Calculate the wavelength inside the glass.

Use the average refractive index:

\[ \lambda_{\text{glass}}=\frac{\lambda_0}{n_{\text{glass}}} \]

Substitute:

\[ \begin{aligned} \lambda_{\text{glass}} &=\frac{589}{1.499}\\ &=392.9\text{ nm} \end{aligned} \]

Therefore:

\[ \boxed{\lambda_{\text{glass}}\approx393\text{ nm}} \]

(d) Explain the experimental limitations.

Repeating measurements reduces the influence of random errors in reading the protractor and identifying the exact ray position.

If the group used only one trial, an inaccurate angle measurement could produce an inaccurate refractive index. Multiple trials allow the group to calculate an average and identify inconsistent readings.

There is also an important issue with the emergent ray.

When light passes through a rectangular glass block with parallel faces, it refracts at the first boundary and again at the second boundary. The emerging ray is parallel to the original incident ray, although it is laterally displaced.

Therefore, the angle of emergence is not the angle of refraction inside the glass.

The group must trace the internal ray segment and measure its angle at the first boundary. Alternatively, a suitable semicircular block arrangement can isolate refraction at one flat boundary.

Question 5

A monochromatic laser has a vacuum wavelength of 600 nm.

It travels through water, then passes through two different glass layers before returning to water.

The arrangement is:

\[ \text{Water}\rightarrow \text{Glass A}\rightarrow \text{Glass B}\rightarrow \text{Water} \]

The following illustrative values are supplied:

All four interfaces are parallel, and the laser strikes the first boundary at \(40^\circ\) to the normal.

Assume the materials are transparent and the refractive indices apply to the laser’s wavelength.

(a) Calculate the wavelength in water, Glass A, and Glass B.

(b) Calculate the angle of the ray inside Glass A and Glass B.

(c) Determine the angle of the ray after it emerges back into water.

(d) A student starts with the wavelength in water and divides it by the absolute refractive index of Glass A:

\[ \frac{451}{1.50}\approx301\text{ nm} \]

Explain why this calculation is incorrect and determine the correct wavelength in Glass A.

(e) A detector measures the wavelength and direction of the laser before and after the entire arrangement. Both measurements give approximately 451 nm and \(40^\circ\).

A student concludes that the wavelength never changed inside either glass layer.

Is this conclusion justified? Explain what the external measurements do and do not tell us.

Solution 5

The wavelengths are approximately 451 nm in water, 400 nm in Glass A, and 375 nm in Glass B. The ray travels at approximately \(34.7^\circ\) in Glass A and \(32.3^\circ\) in Glass B, before emerging into water at \(40^\circ\).

The student’s conclusion is incorrect. The wavelength changes inside each glass layer, even though the outgoing wavelength and direction return to their original values.

(a) Calculate the wavelength in each material.

The vacuum wavelength is:

\[ \lambda_0=600\text{ nm} \]

For water:

\[ \begin{aligned} \lambda_{\text{water}} &=\frac{600}{1.33}\\ &=451.1\text{ nm} \end{aligned} \]

For Glass A:

\[ \begin{aligned} \lambda_A &=\frac{600}{1.50}\\ &=400\text{ nm} \end{aligned} \]

For Glass B:

\[ \begin{aligned} \lambda_B &=\frac{600}{1.60}\\ &=375\text{ nm} \end{aligned} \]

Therefore:

When the light returns to water, its wavelength returns to approximately 451.1 nm.

(b) Calculate the angle inside Glass A.

The ray travels from water into Glass A.

Apply Snell’s law:

\[ n_{\text{water}}\sin i =n_A\sin\theta_A \]

Rearrange:

\[ \sin\theta_A =\frac{n_{\text{water}}}{n_A}\sin i \]

Substitute:

\[ \begin{aligned} \sin\theta_A &=\frac{1.33}{1.50}\sin40^\circ\\ &=0.570 \end{aligned} \]

Take the inverse sine:

\[ \boxed{\theta_A\approx34.7^\circ} \]

Now calculate the angle inside Glass B.

At the Glass A-Glass B boundary, the ray travels from a refractive index of 1.50 into a refractive index of 1.60.

Snell’s law gives:

\[ n_A\sin\theta_A=n_B\sin\theta_B \]

Substitute:

\[ \begin{aligned} \sin\theta_B &=\frac{1.50}{1.60}\sin34.7^\circ\\ &\approx0.534 \end{aligned} \]

Therefore:

\[ \boxed{\theta_B\approx32.3^\circ} \]

The ray bends closer to the normal when it enters Glass B because its refractive index is higher than that of Glass A.

(c) Calculate the final angle in water.

The ray now travels from Glass B back into water.

Apply Snell’s law:

\[ n_B\sin\theta_B =n_{\text{water}}\sin\theta_{\text{out}} \]

Substitute:

\[ \begin{aligned} \sin\theta_{\text{out}} &=\frac{1.60}{1.33}\sin32.3^\circ\\ &\approx0.643 \end{aligned} \]

Taking the inverse sine:

\[ \boxed{\theta_{\text{out}}\approx40^\circ} \]

The ray emerges parallel to its original direction.

This happens because all the interfaces are parallel and the initial and final media are both water. The ray may still be laterally displaced.

(d) Explain the incorrect wavelength calculation.

The student has mixed an absolute refractive index with a wavelength measured in a different medium.

The value 1.50 is the absolute refractive index of Glass A, relative to vacuum.

The value 451 nm, however, is the wavelength in water, not vacuum.

To move directly from water into Glass A, we need the relative refractive index:

\[ n_{A/\text{water}} =\frac{n_A}{n_{\text{water}}} \]

Substitute:

\[ \begin{aligned} n_{A/\text{water}} &=\frac{1.50}{1.33}\\ &=1.128 \end{aligned} \]

Now calculate:

\[ \begin{aligned} \lambda_A &=\frac{\lambda_{\text{water}}}{n_{A/\text{water}}}\\ &=\frac{451.1}{1.128}\\ &\approx400\text{ nm} \end{aligned} \]

Therefore:

\[ \boxed{\lambda_A=400\text{ nm}} \]

The student’s mistake was using the absolute refractive index of glass as though it were the relative refractive index between water and glass.

(e) Evaluate the student’s conclusion.

The conclusion is not justified.

The wavelength and direction measured outside the arrangement are the same as those measured before the light entered because the light starts and finishes in the same medium, and all the interfaces are parallel.

However, the internal wavelengths were different:

• Water: 451.1 nm.
• Glass A: 400 nm.
• Glass B: 375 nm.

The external measurements do not reveal these intermediate changes.

This is the important distinction: the final wavelength and direction depend on the initial and final media, but the light’s speed and wavelength inside the arrangement depend on each material it passes through.

A measurement of accumulated optical phase using interference could reveal the delay introduced by the glass layers, even when the outgoing wavelength and direction are unchanged.

This is why identifying the medium pair at every boundary matters. Once you can track the wavelength through several materials, the next decision is whether the ray can refract at each boundary or undergoes total internal reflection.

For that next step, see How to Decide if a Ray Refracts or Reflects at a Second Boundary.