Consider sunlight incident on a slit of width 10⁴AA. The image seen through the slit shall (a) be a fine sharp

NCERT Exemplar · Q1

10^4\ \text{\AA}

(a) be a fine sharp slit white in colour at the center.

(b) a bright slit white at the center diffusing to zero intensities at the edges.

(d) only be a diffused slit white in colour.

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Concept understanding — Frequency Invariance

Frequency Invariance

When a light wave crosses from one medium into another — on reflection or on refraction — one property never changes: its frequency. Everything else about the wave (its speed, its wavelength) can change, but the frequency is fixed the moment the wave leaves its source.

Why frequency is set by the source, not the medium

A wave's frequency is the rate at which its source oscillates. Think of shaking one end of a rope: if you shake it 5 times a second, exactly 5 crests leave your hand every second. If that rope changes into a heavier rope partway along, the wave travels slower in the heavier section, but the number of crests arriving per second at the join must still equal 5 — a crest cannot be created or destroyed at the boundary. The same logic applies to light: whatever surface it meets, the boundary condition (continuity of the oscillating electric and magnetic fields) forces the reflected and refracted waves to oscillate at exactly the incident frequency.

Watch out

A common mistake is to think that because wavelength changes across a boundary, frequency must change too. It's the reverse: frequency is fixed by the source, so when speed changes, wavelength ( $\lambda = v/f$ ) adjusts to compensate.

What changes instead: speed and wavelength

In a medium of refractive index $n$ , light slows to $v = c/n$ . Since frequency $f$ is unchanged and $v = f\lambda$ , the wavelength inside the medium must shrink:

f_{\text{medium}} = f_{\text{vacuum}}, \qquad v = \frac{c}{n}, \qquad \lambda_{\text{medium}} = \frac{\lambda_{\text{vacuum}}}{n}

For reflection, the ray stays in the original medium, so speed, wavelength, and frequency are all unchanged. For refraction, the frequency still matches the incident wave, but speed and wavelength both scale by $1/n$ .

Does slowing down mean losing energy?

No. The energy of light is carried by its photons, each of energy $E = hf$ — a quantity that depends only on frequency. Since frequency doesn't change on entering a denser medium, the energy per photon is unchanged too; only the wave's speed and wavelength are affected. (The wave's amplitude does adjust at the boundary so that energy is properly split between the reflected and transmitted beams — but frequency, and hence photon energy, is untouched.)

Worked example

Light of $\lambda_0 = 589\ \text{nm}$ in air strikes water ( $n = 1.33$ ). The frequency is

f = \frac{c}{\lambda_0} = \frac{3\times10^8}{589\times10^{-9}} \approx 5.09\times10^{14}\ \text{Hz}

This value is the same for the reflected ray (still in air, $\lambda = 589\ \text{nm}$ ) and the refracted ray (now in water, where $\lambda' = \lambda_0/n \approx 443\ \text{nm}$ and $v' = c/n \approx 2.26\times10^8\ \text{m/s}$ ).

The takeaway

Frequency is the one wave property that survives reflection and refraction unchanged, because it is fixed by the source and enforced by the boundary condition at every interface. Speed and wavelength are the properties that adapt to the medium.

Frequency invariance across reflection and refraction is a core idea in the NCERT/CBSE Class 12 Physics chapter on Wave Optics, and students searching for "why does frequency not change in refraction" or preparing "wave optics important questions" for JEE Main and NEET will find this exact reasoning tested repeatedly. Understanding this distinction between frequency, wavelength, and speed is also a favourite conceptual trap in board-exam and competitive-exam MCQs on light.

Why this formula?

Frequency Invariance

When light (or any wave) crosses from one medium into another, one property refuses to change: its frequency. Understanding why is the key to Snell's law and to how colour is preserved through glass, water and lenses.

On refraction the frequency $f$ stays the same; the speed $v$ and wavelength $\lambda$ change together so that $v = f\lambda$ still holds.

Why Frequency Is Conserved

A wave is driven at the boundary by the incoming oscillation. The electric field of the light wave forces the electrons in the second medium to oscillate, and they can only oscillate at the same rate at which they are driven. If the frequency changed, wave crests would either pile up at or vanish from the interface — the boundary would not stay continuous. So the number of crests arriving per second must equal the number leaving per second:

f_1 = f_2 = f

What Does Change

Inside a denser medium light slows to $v = c/n$ . Since $f$ is fixed and $v = f\lambda$ , the wavelength must shrink in the same proportion:

\lambda_{\text{medium}} = \frac{v}{f} = \frac{c/n}{f} = \frac{\lambda_{\text{vacuum}}}{n}

So in glass of $n=1.5$ , both speed and wavelength fall to two-thirds of their vacuum values, but the frequency — and therefore the colour — is unchanged.

Worked Idea

Red light, $\lambda_0 = 660\ \text{nm}$ in air, enters water ( $n = 1.33$ ).

Frequency: $f = c/\lambda_0 = \dfrac{3\times10^{8}}{660\times10^{-9}} \approx 4.5\times10^{14}\ \text{Hz}$ — unchanged in water.
Wavelength in water: $\lambda = 660/1.33 \approx 496\ \text{nm}$ .

Note

This is why an object under water keeps its colour: our eyes respond to frequency, and frequency is the invariant quantity.

$a = 10^4\ \text{\AA} = 1\ \mu\text{m}$ is comparable to the visible wavelength, so the image is a sharp white central strip with the colours diffusing at the ends — matching option (c).

1. Set up the numbers.

a = 10^4\ \text{\AA} = 10^4 \times 10^{-10}\ \text{m} = 10^{-6}\ \text{m} = 1\ \mu\text{m}.

Visible light spans $\lambda \approx 4000\ \text{\AA}$ (violet) to $7000\ \text{\AA}$ (red), i.e. $400$ – $700\ \text{nm}$ .

2. Compare with the wavelength. The ratio $a/\lambda$ ranges from $1000/700 \approx 1.4$ to $1000/400 = 2.5$ . The slit is only a small multiple of the wavelength, so diffraction is moderate — clearly present but not overwhelming.

3. Diffraction angle. The angular half-width of the central maximum is $\theta \approx \lambda/a$ , which is larger for red than for violet:

\theta_{\text{red}} \approx \frac{700}{1000} = 0.7\ \text{rad}, \qquad \theta_{\text{violet}} \approx \frac{400}{1000} = 0.4\ \text{rad}.

4. What appears on the screen. At the centre ( $\theta = 0$ ) every wavelength has its principal maximum, so all colours overlap and add to give white. Moving outward, the maxima of the different colours fall at different angles (red spreads more than violet), so the edges of the image show the colours fanned out — this matches option (c): a bright white centre diffusing to regions of different colours.

✓Final answer

Option (c). The image is a fine, sharp white slit at the centre, with the colours diffusing (spreading) towards the edges.

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set 55/1/11 markMCQ
$\left(\dfrac{3}{2}\right)$
›Reveal solutionSolution
The key idea is that frequency is determined by the source and never changes when a wave enters a new medium, while wavelength scales with the wave speed. Here the speed reduces by a factor of 2, so the wavelength is halved — making option (C) correct.

When a wave crosses from one medium into another, the frequency never changes — it is set by the source and cannot be altered by the medium. What does change is the wave speed, and with it the wavelength, because $v = f \lambda$ must hold in every medium.

The question gives us the relative permittivity $\varepsilon_r = \frac{3}{2}$ and relative permeability $\mu_r = \frac{8}{3}$ . These determine the refractive index of the dielectric, which tells us how much the speed changes.

Find the refractive index. For any medium, the refractive index is $n = \sqrt{\varepsilon_r \mu_r}$ (for a non-magnetic medium $\mu_r \approx 1$ and this reduces to $n = \sqrt{\varepsilon_r}$ ; here $\mu_r = \frac{8}{3}$ , so we keep both factors).

$n = \sqrt{\frac{3}{2} \times \frac{8}{3}} = \sqrt{\frac{8}{2}} = \sqrt{4} = 2.$
So the dielectric has refractive index $n = 2$ .

Relate speed to wavelength. In vacuum, speed is $c$ , wavelength is $\lambda_0$ , and $c = f \lambda_0$ . In the medium, speed is $v = \frac{c}{n} = \frac{c}{2}$ , and $v = f \lambda$ . Since $f$ is unchanged,

$\frac{c}{2} = f \lambda \quad \Rightarrow \quad \lambda = \frac{c}{2f} = \frac{\lambda_0}{2}.$
The wavelength is halved.

Check the options.

(A) says wavelength doubled — wrong.

(B) says wavelength doubled and frequency halved — both wrong.

(C) says wavelength halved, frequency unchanged — matches our result.

(D) says both unchanged — wrong.

Watch out

A common mistake is to forget that frequency is invariant across media. Many students incorrectly apply $v = f \lambda$ as if $f$ could change, leading them to pick (B) or (D). Always remember: frequency is a source property, not a medium property.

Tip

You can also think of this as: $n = \frac{\lambda_0}{\lambda}$ directly, since $n = \frac{c}{v} = \frac{f \lambda_0}{f \lambda} = \frac{\lambda_0}{\lambda}$ . With $n=2$ , $\lambda = \lambda_0/2$ in one step.

✓Final answer
$The correct option is (C) .$
CBSE 2026Set ANNUAL1 markMCQ
$n_1$
(i) Frequency of incident, refracted and reflected light is same.
(ii) Frequency of incident and reflected light is same but frequency of refracted wave is decreased.
(iii) Frequency of incident and reflected light is same but frequency of refracted light is increased.
(iv) Frequency of incident light and refracted light is same but frequency of reflected light is changed.
›Reveal solutionSolution

Frequency of light is fixed by the source and never changes on reflection or refraction.

The frequency of a light wave is determined by the source that produces it, not by the medium it travels through. On reflection and refraction only the wavelength and speed change (since $v=f\lambda$ and $v$ depends on the medium); the frequency of the incident, reflected and refracted light all remain the same.

✓Final answer
$(i) Frequency of incident, refracted and reflected light is same.$
CBSE 2026Set ANNUAL1 markMCQ
$Q.If a wave gets refracted into a denser medium, then which of the following is true?$
(a) wavelength, speed and frequency decrease.
(b) wavelength increases, speed decreases and frequency remain constant.
(c) wavelength and speed decrease but frequency remains constant.
(d) wavelength, speed and frequency increase.
›Reveal solutionSolution
Frequency is fixed by the source and never changes on refraction; since $v = f\lambda$ and $v$ decreases in a denser medium, $\lambda$ must decrease too.

When a wave crosses from a rarer to a denser medium:

Frequency is determined only by the source that generates the wave and stays the same in every medium — this is why frequency is used to define colour of light, unlike wavelength.

Speed decreases in a denser medium because the refractive index $n = c/v$ is larger there, so $v = c/n$ is smaller.

Wavelength follows from $v = f\lambda \Rightarrow \lambda = v/f$ . Since $f$ is constant and $v$ decreases, $\lambda$ must also decrease proportionally.

This is exactly what Huygens' construction predicts: the wavefronts get compressed (closer together) as they slow down while crossing into the denser medium.

✓Final answer
$(c) Wavelength and speed decrease, but frequency remains constant.$
CBSE 2024Set IMPROVEMENT1 markMCQ
$Q. When light enters from one medium to other medium then its —$
(a) Frequency is changed, wavelength remains unchanged.
(b) Frequency remains unchanged, wavelength is changed.
(c) Frequency and wavelength both are changed.
(d) Frequency and wavelength both remain unchanged.
›Reveal solutionSolution

On crossing a boundary between media, frequency (fixed by the source) stays the same; wavelength changes because speed changes.

When light passes from one medium to another, its frequency $\nu$ is determined by the source and does not change. However, its speed $v$ changes because it depends on the refractive index of the medium ( $v=c/n$ ). Since $v=\nu\lambda$ , a change in speed at constant frequency means the wavelength $\lambda$ must change (it decreases when light enters a denser medium and increases when it enters a rarer medium).

✓Final answer
$(b) Frequency remains unchanged, wavelength is changed.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. For a wave propagating in a medium, identify the property that is independent of the other :$
(a) velocity
(b) wavelength
(c) frequency
(d) All these depends on each other
›Reveal solutionSolution
Frequency is fixed by the source and stays unchanged as a wave crosses into a new medium; velocity and wavelength both change together via v = fλ.

For a wave $v = f\lambda$ :

Velocity (v) depends on the properties of the medium (elasticity/density for mechanical waves, permittivity/permeability for EM waves) — it changes when the wave enters a new medium.

Wavelength (λ) depends on both the medium (through v) and the frequency, since $\lambda = v/f$ — so it changes whenever v changes.

Frequency (f) is determined solely by the source producing the wave (the rate of oscillation) and does not change when the wave passes from one medium to another; only v and λ adjust to keep $v=f\lambda$ satisfied.

So frequency is the one property that does not depend on / is not altered by the medium.

✓Final answer
$Frequency — option (c).$
CBSE 2024Set ANNUAL1 mark
$Q. A monochromatic light travels from a rarer to a denser medium. Does the decrease in speed imply a reduction in the energy carried by a light wave?$
›Reveal solutionSolution

Frequency is fixed by the source and unaffected by the medium; energy per photon depends only on frequency.

When monochromatic light travels from a rarer to a denser medium, its speed decreases ( $v=c/n$ ) and its wavelength decreases proportionally ( $\lambda = \lambda_0/n$ ), but its frequency stays exactly the same, since frequency is determined by the source of the light, not by the medium through which it travels. Since the energy carried by each photon of the wave is $E=h\nu$ , and $\nu$ is unchanged, the energy per photon (and hence the energy carried by the wave, apart from a small loss to partial reflection at the interface) is not reduced by the decrease in speed.

✓Final answer
$E=h\nu$
CBSE 2023Set MODEL1 markMCQ
$Q. When a ray of light enters a glass slab from air, then its:$
(a) Wavelength increases
(b) Wavelength decreases
(c) Frequency increases
(d) Frequency decreases
›Reveal solutionSolution

Frequency stays fixed on refraction; since glass is optically denser, speed and hence wavelength decrease.

When light passes from air into glass, its frequency $\nu$ does not change (it is fixed by the source). The speed of light decreases ( $v = c/n$ , and $n_{glass}>1$ ), and since $v=\nu\lambda$ , the wavelength $\lambda = v/\nu$ decreases as the ray enters the denser medium.

✓Final answer
$(b) Wavelength decreases$
CBSE 2023Set ANNUAL1 markMCQ
$Q.When sound wave is refracted from air to water which of the following quantities remains unchanged?$
(1) Wavelength
(2) Wave number
(3) Wave velocity
(4) Frequency
›Reveal solutionSolution
Frequency of a wave is determined by the source producing it and does not change when the wave crosses into a different medium - only wave speed and wavelength change (since v = f lambda, with f fixed).

When a sound wave refracts from air into water:

The speed of sound changes (it is faster in water, ~1480 m/s, than in air, ~340 m/s), because speed depends on the elastic and inertial properties of the medium (bulk modulus and density).

Since v = f lambda and v changes while f is fixed, the wavelength must also change (it increases when the wave enters the faster medium, water).

The wave number k = 2 pi/lambda is inversely related to wavelength, so it changes too.

The frequency of the wave, however, is set by the source of the vibration (how many oscillations per second it produces) and is a boundary condition that stays the same across the interface - every particle at the boundary, and hence in either medium, oscillates at the same rate. So frequency alone remains unchanged.

✓Final answer
$(4) Frequency.$
CBSE 2020Set NC1 markMCQ
$Q. When light travels from an optically rarer medium to an optically denser medium, the velocity of light decreases because of change in$
(a) frequency
(b) wavelength
(c) amplitude
(d) phase
›Reveal solutionSolution

Frequency is fixed by the source and can't change on crossing a boundary; since $v=f\lambda$ and $v$ falls in the denser medium, it must be $\lambda$ that shrinks.

Reasoning

When light passes from one medium to another, its frequency remains unchanged (it is set by the source and must match on both sides to keep the wave continuous at the boundary). Since

$v = f\lambda$

and $v$ (speed) decreases on entering the optically denser medium while $f$ stays fixed, $\lambda$ (wavelength) must decrease proportionally.

✓Final answer
$The change is in wavelength — option (b).$
CBSE 2018Set ANNUAL1 markMCQ
$Q. When light wave travels from air to glass$
(a) its wavelength decreases
(b) its wavelength increases
(c) its wavelength remains unchanged
(d) its frequency decreases
›Reveal solutionSolution
When light crosses from a rarer to a denser medium its frequency is unchanged (set by the source), but its speed falls, so its wavelength must also fall.

When a light wave travels from air into glass:

The frequency v of the wave is determined by the source and does NOT change on refraction — it stays the same in both media.

The speed of light in glass (v_glass = c/n, where n > 1 is glass's refractive index) is LESS than its speed in air.

Since wave speed v = v (frequency) x lambda (wavelength), and v decreases while the frequency stays fixed, the wavelength lambda = v/(frequency) must decrease.

So the wave 'compresses' — same frequency, shorter wavelength, slower speed — as it enters the optically denser medium.

✓Final answer
$(a) its wavelength decreases.$

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Q.Consider sunlight incident on a slit of width $10^4\ \text{\AA}$ . The image seen through the slit shall

Concept understanding — Frequency Invariance

Frequency Invariance

Why frequency is set by the source, not the medium

What changes instead: speed and wavelength

Does slowing down mean losing energy?

Worked example

The takeaway

Why this formula?

Frequency Invariance

Why Frequency Is Conserved

What Does Change

Worked Idea

Method: Comparing Aperture Size to Wavelength to Predict the Diffraction Pattern

Steps

Q.Consider sunlight incident on a slit of width 104 A˚10^4\ \text{\AA}104 A˚. The image seen through the slit shall

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Q.Consider sunlight incident on a slit of width $10^4\ \text{\AA}$ . The image seen through the slit shall

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