Why Light’s Frequency Stays Constant When It Enters Glass

You see an exam question where light enters glass, slows down, and has a shorter wavelength. You write \(v=f\lambda\), then conclude that its frequency must decrease as well. The equation is correct. Your conclusion isn’t.

The mistake is assuming that because the speed changes, the frequency must change too. For ordinary refraction at a stationary boundary, the frequency stays constant while the speed and wavelength change.

Before reading on, make a prediction. Suppose light enters glass and its speed drops to two-thirds of its original value. What happens to its frequency and wavelength?

The frequency stays the same. The wavelength becomes two-thirds of its original value. The key is understanding what the source controls and what the medium changes.

01Diagnose the mistake: speed, frequency, and wavelength are different things

Three quantities describe a wave:

• Frequency, \(f\): the number of complete oscillations passing a fixed point each second, measured in hertz (Hz).
• Wavelength, \(\lambda\): the distance between consecutive points in the same phase, such as two neighbouring crests, measured in metres (m).
• Wave speed, \(v\): the distance the wave pattern travels each second, measured in metres per second (m/s).

These quantities are connected, but they do not have to change together.

Imagine a conveyor belt carrying evenly spaced marks past a sensor. The rate at which the marks arrive tells you their frequency, and the spacing between marks tells you their wavelength. If the belt slows down while the supply rate stays constant, the marks become closer together.

The analogy captures the relationship between speed, spacing, and arrival rate. Light is not made of little marks on a belt, but the same mathematical relationship applies.

For light, the source determines the frequency. When the light enters a different medium, its speed changes. The wavelength adjusts to keep the frequency unchanged.

The common mistake is to treat \(f\) and \(\lambda\) as fixed while changing \(v\). That is not what happens during ordinary refraction. At a stationary boundary, the transmitted light oscillates at the same frequency as the incident light.

02Use the correct wave equation for each medium

The wave equation is

\[ v=f\lambda \]

It applies to waves in different media, including light. The important detail is that you must use the speed and wavelength for the same medium.

In a vacuum, the speed of light is \(c\), approximately \(3.00\times10^8\ \text{m/s}\). The equation becomes

\[ c=f\lambda_0 \]

Here, \(\lambda_0\) is the wavelength in a vacuum.

Inside glass, the speed is lower than in a vacuum, so we use

\[ v=f\lambda_g \]

Here, \(v\) is the speed in glass and \(\lambda_g\) is the wavelength in glass. Air is often approximated as a vacuum in HSC calculations.

The refractive index, \(n\), describes how much the speed of light decreases in a medium:

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

Rearranging gives

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

For glass with \(n=1.50\),

\[ v=\frac{3.00\times10^8}{1.50} \]

\[ v=2.00\times10^8\ \text{m/s} \]

The light travels at two-thirds of its vacuum speed.

Now compare the two wavelength equations. Because the frequency remains the same,

\[ \lambda_0=\frac{c}{f} \]

and

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

Substituting \(v=c/n\) gives the crucial result:

\[ \boxed{\lambda_g=\frac{\lambda_0}{n}} \]

For this glass, the wavelength becomes two-thirds of its vacuum value. The frequency does not change.

03How this connects to refraction

The change in speed and wavelength also explains why light bends when it enters glass at an angle.

A wavefront joins points on a wave that are at the same stage of oscillation. The perpendicular direction to a wavefront is the direction in which the wave propagates. A ray represents that direction of propagation.

When light enters glass obliquely, one side of a wavefront reaches the boundary before the other side. The first part slows down in the glass while the remaining part is still travelling through air. This uneven change in speed changes the wavefront’s orientation, so the direction of propagation changes.

The important point is that the direction changes because the wave speed changes across the boundary, not because the frequency changes.

For light entering glass from air at an angle, the ray bends towards the normal. The normal is an imaginary line perpendicular to the boundary.

There are two useful cases to distinguish:

• Normal incidence: The light enters along the normal. Its speed and wavelength change, but its direction does not.
• Oblique incidence: The light enters at an angle to the normal. Its speed and wavelength change, and its direction bends towards the normal when entering glass from air.

If light travels from glass into air, the process reverses: its speed and wavelength increase, its frequency remains unchanged, and the ray bends away from the normal.

The angle changes are described by Snell’s law:

\[ n_1\sin\theta_1=n_2\sin\theta_2 \]

Here, \(n_1\) and \(n_2\) are the refractive indices of the two media, and \(\theta_1\) and \(\theta_2\) are the angles measured from the normal.

Snell’s law tells you how the direction changes. The wave equation tells you how speed, frequency, and wavelength are related. Use both when a question asks about wavelength and refraction.

04The exam fix: a reliable decision method

When you see a question about light entering a new medium, use this sequence.

1. Identify the medium. Check whether each speed or wavelength refers to air, a vacuum, or glass.
2. Write the appropriate wave equation. Use \(c=f\lambda_0\) in a vacuum, or \(v=f\lambda\) in a medium.
3. Keep the frequency unchanged. For ordinary refraction at a stationary boundary, the transmitted light has the same frequency as the incident light.
4. Calculate the speed if necessary. If the refractive index is given, use \(v=c/n\).
5. Calculate the wavelength. Use \(\lambda=v/f\), with the speed and wavelength referring to the same medium.
6. Check the physical result. If light enters glass from air, its speed and wavelength decrease. If it leaves glass for air, both increase. Its frequency stays the same in either direction.

A useful final check is this:

\[ \frac{\lambda_2}{\lambda_1} =\frac{v_2}{v_1} \qquad\text{when }f\text{ is unchanged} \]

The ratio of wavelengths must equal the ratio of speeds. If the speed halves, the wavelength halves. If the speed doubles, the wavelength doubles.

Wording changes that catch students out

One qualification matters: frequency remains unchanged for ordinary refraction at a stationary boundary. A moving source or moving boundary can produce a Doppler frequency shift, which is a different situation.

05Questions and solutions

Question 1

Monochromatic light enters glass at normal incidence. Its wavelength in air is \(600\ \text{nm}\). The speed of light in glass is \(2.0\times10^8\ \text{m/s}\), compared with \(3.0\times10^8\ \text{m/s}\) in air.

Calculate its wavelength in glass and state whether its direction changes.

Solution 1

The wavelength becomes \(400\ \text{nm}\), and the direction does not change. The frequency remains constant at the stationary boundary, so the wavelength changes in the same ratio as the speed.

\[ \begin{aligned} \lambda_g&=\lambda_{\text{air}}\frac{v_g}{v_{\text{air}}}\\[4pt] &=600\ \text{nm}\times\frac{2.0\times10^8}{3.0\times10^8}\\[4pt] &=400\ \text{nm} \end{aligned} \]

The wavelength decreases because the light travels more slowly in glass. Since the light enters at normal incidence, all parts of the wavefront cross the boundary in the same way, so there is no change in direction.

Question 2

Inside a transparent block, monochromatic light has a speed of \(1.80\times10^8\ \text{m/s}\) and a wavelength of \(360\ \text{nm}\).

Calculate:

(a) the refractive index of the block;

(b) the frequency of the light;

(c) the wavelength of the light in a vacuum.

Take \(c=3.00\times10^8\ \text{m/s}\).

Solution 2

The refractive index is \(1.67\), the frequency is \(5.00\times10^{14}\ \text{Hz}\), and the vacuum wavelength is \(600\ \text{nm}\). Work backwards from the speed and wavelength in the block, then use the unchanged frequency to find the vacuum wavelength.

(a) Refractive index

\[ \begin{aligned} n&=\frac{c}{v}\\[4pt] &=\frac{3.00\times10^8}{1.80\times10^8}\\[4pt] &=1.67 \end{aligned} \]

(b) Frequency

Convert the wavelength to metres:

\[ 360\ \text{nm}=360\times10^{-9}\ \text{m} \]

Then use the wave equation for the medium.

\[ \begin{aligned} f&=\frac{v}{\lambda}\\[4pt] &=\frac{1.80\times10^8}{360\times10^{-9}}\\[4pt] &=5.00\times10^{14}\ \text{Hz} \end{aligned} \]

(c) Vacuum wavelength

The frequency is unchanged when the light moves between the block and a vacuum.

\[ \begin{aligned} \lambda_0&=\frac{c}{f}\\[4pt] &=\frac{3.00\times10^8}{5.00\times10^{14}}\\[4pt] &=6.00\times10^{-7}\ \text{m}\\[4pt] &=600\ \text{nm} \end{aligned} \]

The vacuum wavelength is longer than the wavelength in the block. Their ratio, \(600/360=1.67\), agrees with the calculated refractive index. This independent check confirms that the values are consistent.

Question 3

Light travels through glass with a refractive index of \(1.60\) and a wavelength of \(400\ \text{nm}\). It reaches a glass-air boundary at an angle of \(30^\circ\) to the normal.

Assuming the refractive index of air is \(1.00\), calculate:

(a) the wavelength in air;

(b) the angle of refraction in air;

(c) whether the frequency changes.

Solution 3

The wavelength in air is \(640\ \text{nm}\), the refracted angle is approximately \(53.1^\circ\), and the frequency remains unchanged. Light speeds up as it enters air, so its wavelength increases and its ray bends away from the normal.

(a) Wavelength in air

First calculate the speed in glass:

\[ v_g=\frac{3.00\times10^8}{1.60} =1.875\times10^8\ \text{m/s} \]

Then calculate the frequency in glass:

\[ \begin{aligned} f&=\frac{v_g}{\lambda_g}\\[4pt] &=\frac{1.875\times10^8}{400\times10^{-9}}\\[4pt] &=4.6875\times10^{14}\ \text{Hz} \end{aligned} \]

Use the same frequency to calculate the wavelength in air:

\[ \begin{aligned} \lambda_{\text{air}}&=\frac{c}{f}\\[4pt] &=\frac{3.00\times10^8}{4.6875\times10^{14}}\\[4pt] &=640\times10^{-9}\ \text{m}\\[4pt] &=640\ \text{nm} \end{aligned} \]

The wavelength increases from \(400\ \text{nm}\) to \(640\ \text{nm}\).

(b) Angle of refraction

Apply Snell’s law:

\[ n_g\sin\theta_g=n_{\text{air}}\sin\theta_{\text{air}} \]

Substitute the values:

\[ 1.60\sin30^\circ=1.00\sin\theta_{\text{air}} \]

\[ \sin\theta_{\text{air}}=0.80 \]

\[ \theta_{\text{air}}\approx53.1^\circ \]

The refracted angle is larger than the incident angle, so the ray bends away from the normal.

(c) Frequency

The frequency remains \(4.6875\times10^{14}\ \text{Hz}\). The change in wavelength follows from the increase in speed; the wave does not need a new frequency to travel faster.

Question 4

Red light has a wavelength of \(660\ \text{nm}\) in air, while blue light has a wavelength of \(440\ \text{nm}\) in air. Both beams enter the same glass at an incident angle of \(45^\circ\).

For this glass, the refractive index is \(1.50\) for red light and \(1.53\) for blue light.

A student claims:

“Blue light bends more because its frequency increases more when it enters glass.”

Explain the error in this claim. Calculate the wavelength of each colour in glass and determine the refracted angle for each beam.

Solution 4

The student’s explanation is incorrect: neither frequency increases on entering glass. Blue light bends slightly more because its refractive index in this glass is higher than that of red light. The two colours already have different frequencies in air, and each retains its own frequency at the stationary boundary.

Step 1: Calculate the wavelengths in glass.

For red light,

\[ \begin{aligned} \lambda_r&=\frac{\lambda_{\text{air},r}}{n_r}\\[4pt] &=\frac{660}{1.50}\\[4pt] &=440\ \text{nm} \end{aligned} \]

For blue light,

\[ \begin{aligned} \lambda_b&=\frac{\lambda_{\text{air},b}}{n_b}\\[4pt] &=\frac{440}{1.53}\\[4pt] &\approx288\ \text{nm} \end{aligned} \]

Blue light has the shorter wavelength in glass. This follows from its shorter starting wavelength and the different refractive index for that colour.

Step 2: Calculate the refracted angles.

For red light, Snell’s law gives

\[ \sin\theta_r=\frac{\sin45^\circ}{1.50} \]

\[ \theta_r\approx28.1^\circ \]

For blue light,

\[ \sin\theta_b=\frac{\sin45^\circ}{1.53} \]

\[ \theta_b\approx27.5^\circ \]

Blue light refracts to a slightly smaller angle from the normal, so it bends more towards the normal.

Step 3: Check the frequency claim.

The frequencies in air are approximately

\[ f_r=\frac{3.00\times10^8}{660\times10^{-9}} =4.55\times10^{14}\ \text{Hz} \]

and

\[ f_b=\frac{3.00\times10^8}{440\times10^{-9}} =6.82\times10^{14}\ \text{Hz} \]

These frequencies remain unchanged in the glass. Blue light has the higher frequency before it reaches the boundary; it does not acquire a higher frequency because it bends more.

This is an example of dispersion: the refractive index of a material can depend on the frequency of the light. That changes the speeds, wavelengths, and refraction angles of different colours without changing their individual frequencies.

Question 5

An ideal, non-dispersive glass slab has a thickness of \(0.60\ \text{m}\) and a refractive index of \(1.50\). It is surrounded by air.

A narrow-band light pulse with a carrier frequency of \(5.0\times10^{14}\ \text{Hz}\) contains 12 oscillations. The pulse enters the slab at normal incidence.

Assume negligible reflection and dispersion. Compare the pulse travelling through the glass slab with the same pulse travelling through a \(0.60\ \text{m}\) air gap.

Determine:

(a) the speed and wavelength inside the glass;

(b) the temporal duration of the pulse and its spatial length in air and glass;

(c) how much later the pulse emerges from the glass than it would from the air gap;

(d) whether the shorter wavelength inside glass means the pulse contains fewer oscillations.

Solution 5

The transmitted pulse retains its 12 oscillations and its temporal duration of \(24\ \text{fs}\). Its spatial length decreases from \(7.2\ \mu\text{m}\) in air to \(4.8\ \mu\text{m}\) in glass, and it emerges \(1.0\ \text{ns}\) later than it would through the same thickness of air. The crucial distinction is between the pulse’s length in space and its duration in time.

(a) Speed and wavelength in glass

The speed in glass is

\[ \begin{aligned} v_g&=\frac{c}{n}\\[4pt] &=\frac{3.00\times10^8}{1.50}\\[4pt] &=2.00\times10^8\ \text{m/s} \end{aligned} \]

The wavelength in air is approximately

\[ \lambda_{\text{air}}=\frac{c}{f} =\frac{3.00\times10^8}{5.0\times10^{14}} =600\ \text{nm} \]

The wavelength in glass is

\[ \begin{aligned} \lambda_g&=\frac{v_g}{f}\\[4pt] &=\frac{2.00\times10^8}{5.0\times10^{14}}\\[4pt] &=400\ \text{nm} \end{aligned} \]

The wavelength decreases to two-thirds of its value in air, while the carrier frequency remains unchanged.

(b) Temporal duration and spatial length

The pulse contains 12 oscillations. Its temporal duration is the number of oscillations divided by the frequency:

\[ \begin{aligned} \Delta t&=\frac{N}{f}\\[4pt] &=\frac{12}{5.0\times10^{14}}\\[4pt] &=2.4\times10^{-14}\ \text{s}\\[4pt] &=24\ \text{fs} \end{aligned} \]

Here, \(N\) is the number of oscillations and \(\Delta t\) is the pulse duration.

The spatial length of the pulse is approximately the number of oscillations multiplied by the wavelength.

In air,

\[ L_{\text{air}}=N\lambda_{\text{air}} =12\times600\ \text{nm} =7.2\ \mu\text{m} \]

In glass,

\[ L_g=N\lambda_g =12\times400\ \text{nm} =4.8\ \mu\text{m} \]

The pulse occupies less space in glass, but its temporal duration remains \(24\ \text{fs}\) under the stated non-dispersive assumption.

(c) Extra time to emerge

The time taken to cross a distance is the distance divided by speed.

For the glass slab,

\[ \begin{aligned} t_g&=\frac{0.60}{2.00\times10^8}\\[4pt] &=3.0\times10^{-9}\ \text{s}\\[4pt] &=3.0\ \text{ns} \end{aligned} \]

For the same distance in air,

\[ \begin{aligned} t_{\text{air}}&=\frac{0.60}{3.00\times10^8}\\[4pt] &=2.0\times10^{-9}\ \text{s}\\[4pt] &=2.0\ \text{ns} \end{aligned} \]

Therefore, the extra time is

\[ \begin{aligned} \Delta t_{\text{delay}}&=t_g-t_{\text{air}}\\[4pt] &=3.0-2.0\\[4pt] &=1.0\ \text{ns} \end{aligned} \]

The pulse emerges \(1.0\ \text{ns}\) later than it would through the air gap.

(d) Does the shorter wavelength mean fewer oscillations?

No. The pulse still contains 12 oscillations. The oscillations are packed more closely together in space because the wavelength is shorter.

The tempting mistake is to interpret a shorter spatial length as a shorter duration or fewer oscillations. But spatial length depends on both the number of oscillations and the wavelength:

\[ L=N\lambda \]

The number of oscillations stays the same, so the spatial length decreases when the wavelength decreases. Since the frequency also stays the same, the temporal duration remains unchanged in this ideal non-dispersive model.

This question combines the central ideas: the source determines the frequency, the medium determines the wave speed, the wavelength adjusts to match both, and the reduced speed produces a time delay.

Once you can separate these quantities reliably, the next step in an optics question is to decide whether you need the wave equation, Snell’s law, or both. Use \(v=f\lambda\) to connect speed and wavelength; use Snell’s law to find the change in direction. If several colours are involved, also check whether the refractive index differs with frequency.