Consider sunlight incident on a pinhole of width 10³AA. The image of the pinhole seen on a screen shall be (a)

NCERT Exemplar · Q6

10^3\ \text{\AA}

(a) a sharp white ring.

(b) different from a geometrical image.

(d) diffused coloured region around a sharp central white spot.

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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 …

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:

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

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$a = 10^3\ \text{\AA} = 100\ \text{nm}$ is smaller than the wavelength of visible light, so diffraction is extreme and the screen shows a spread-out, diffuse patch rather than a sharp geometrical image — matching option (b).

1. Size of the pinhole.

a = 10^3\ \text{\AA} = 10^3 \times 10^{-10}\ \text{m} = 10^{-7}\ \text{m} = 100\ \text{nm}.

2. Compare with visible wavelengths. Visible light has $\lambda \approx 400$ – $700\ \text{nm}$ . Here $a < \lambda$ for the whole visible range, i.e. the aperture is sub-wavelength.

3. Diffraction condition. The angular spread of light through an aperture is of order $\theta \sim \lambda/a$ . With $\lambda/a > 1$ for every colour, $\sin\theta$ is of order 1 — the light bends through very large angles. There is no near-parallel beam that could cast a sharp shadow the size of the hole. …

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Q.Consider sunlight incident on a pinhole of width $10^3\ \text{\AA}$ . The image of the pinhole seen on a screen shall be

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

Why this formula?

Frequency Invariance

Why Frequency Is Conserved

What Does Change

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

Steps

Q.Consider sunlight incident on a pinhole of width 103 A˚10^3\ \text{\AA}103 A˚. The image of the pinhole seen on a screen shall be

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Q.Consider sunlight incident on a pinhole of width $10^3\ \text{\AA}$ . The image of the pinhole seen on a screen shall be

Unlock everything free for 14 days