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.

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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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