For light diverging from a point source (a) the wavefront is spherical. (b) the intensity decreases in proportion to the distance squared. (c) the wavefront is parabolic. (d) the intensity at the wavefront does not depend on the distance.

For light diverging from a point source (a) the wavefront is spherical. (b) the intensity decreases in proport

NCERT Exemplar · Q8

Q. For light diverging from a point source

(a) the wavefront is spherical.

(b) the intensity decreases in proportion to the distance squared.

(d) the intensity at the wavefront does not depend on the distance.

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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 point source emits spherical wavefronts, and its intensity falls off as the inverse square of the distance, $I \propto 1/r^2$ — matching options (a) and (b).

1. Shape of the wavefront. A point source radiates symmetrically in every direction. All points reached at the same instant are equidistant from the source, so each surface of constant phase is a sphere centred on the source. Hence the wavefronts are spherical (they look effectively plane only at very large distances). → Option (a) is true.

2. Intensity versus distance. Energy is conserved: the total power $P$ crossing any closed surface around the source is the same. At distance $r$ this power is spread uniformly over a sphere of area $4\pi r^2$ , so

$I(r) = \frac{P}{4\pi r^2} \propto \frac{1}{r^2}.$ …

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Q.For light diverging from a point source

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: Determining Wavefront Shape and Intensity Fall-off From a Point Source

Steps

Q.For light diverging from a point source

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