Q.Monochromatic light of wavelength $589\ \text{nm}$ is incident from air on a water surface. What are the wavelength, frequency and speed of

Concept & Intuition

Ray optics (geometrical optics) treats light as straight lines, ignoring diffraction. But light is a wave, so when it passes through an aperture, it spreads. The question asks: how far must you be from the aperture so that this spreading is negligible — i.e., so that the ray approximation is good?

The answer is the Fresnel distance (or Rayleigh distance) $Z_F \approx a^2 / \lambda$, where $a$ is the aperture size and $\lambda$ the wavelength. Physically, it’s the distance at which the diffraction angle $\theta \approx \lambda / a$ causes a spread just equal to the aperture size itself. Beyond that, the beam diverges significantly; before it, the beam is roughly collimated and ray optics works.

Step-by-step reasoning

1. Identify the relevant formula For a circular aperture of diameter $a$, the Fresnel distance is

This comes from setting the diffraction spread $\approx \theta \cdot Z_F \approx (\lambda / a) \cdot Z_F$ equal to $a$, giving $Z_F = a^2 / \lambda$.

2. Convert all units to a consistent system (metres)

   • Aperture: $a = 0.3\,\text{cm} = 0.3 \times 10^{-2}\,\text{m} = 3 \times 10^{-3}\,\text{m}$.
   • Wavelength: $\lambda = 6000\,\text{Å} = 6000 \times 10^{-10}\,\text{m} = 6 \times 10^{-7}\,\text{m}$.

3. Plug into the formula

4. Interpret the result At distances much less than 15 m, the beam is nearly parallel and ray optics is fine. At distances much greater than 15 m, diffraction dominates and wave optics is needed. So the threshold is exactly 15 m.

The question asks: beyond what distance can we treat light as travelling in straight lines (ray optics) for a given aperture size and wavelength? This is not an arbitrary cutoff — it comes from a fundamental physical condition.

When light passes through an aperture of width $a$, it spreads due to diffraction. The angular spread of the central maximum is roughly $\theta \approx \lambda / a$. Over a distance $L$, this spread widens the beam by an additional amount $L \cdot (\lambda / a)$. Ray optics is a good approximation when this diffraction spread is much smaller than the aperture size itself — that is, when $L \lambda / a \ll a$, or equivalently $L \ll a^2 / \lambda$.

The distance $Z_F = a^2 / \lambda$ is called the Fresnel distance. For $L \ll Z_F$, diffraction is negligible and ray optics works. For $L \gg Z_F$, diffraction dominates and wave optics is needed. The problem asks for the distance at which ray optics becomes a good approximation — that is, the order of $Z_F$ itself.

Let’s compute it step by step.

1. Write the given data in consistent units.

   Aperture $a = 0.3$ cm $= 3 \times 10^{-3}$ m.

   Wavelength $\lambda = 6000$ Å $= 6000 \times 10^{-10}$ m $= 6 \times 10^{-7}$ m.

2. Apply the Fresnel distance formula.

3. Simplify step by step.

Thus, ray optics becomes a good approximation for distances on the order of $15$ m and beyond.

The wave theory of light, championed by Huygens, Fresnel, and Maxwell, treats light as a continuous electromagnetic wave. It beautifully explains phenomena where light bends around obstacles (diffraction), combines to form patterns (interference), and oscillates in a preferred direction (polarization). However, it completely breaks down when explaining how light ejects electrons from a metal surface — the photoelectric effect. The key failure is that wave theory predicts that the energy of ejected electrons should depend on the intensity of light, but experiments show it depends only on the frequency (or color) of light. This puzzle was resolved by Einstein’s photon model, where light behaves as discrete packets of energy.

1. Interference and diffraction are classic wave behaviors. When two waves overlap, they add constructively or destructively (interference). When a wave passes through a slit, it spreads out (diffraction). Both are fully explained by the wave nature of light — no particle picture needed.

2. Polarization is a property unique to transverse waves. Light waves oscillate perpendicular to their direction of travel; polarization filters select waves oscillating in a specific plane. This is perfectly consistent with the wave model.

3. The photoelectric effect is where the wave picture fails. In this effect, light shining on a metal surface ejects electrons. According to wave theory:

   • The energy of the ejected electrons should increase with the intensity (brightness) of light, because a more intense wave carries more energy.
   • Even very dim light should eventually eject electrons if you wait long enough for the wave to transfer enough energy.
   • The effect should occur for any frequency of light, given sufficient intensity.

   But experiments (by Hertz, Lenard, and others) showed the opposite:

   • No electrons are ejected if the light’s frequency is below a certain threshold, no matter how intense the light.
   • Above that threshold, the kinetic energy of ejected electrons depends only on frequency, not intensity.
   • Electrons are ejected instantly, even with very dim light — no time delay.

   These results are impossible to explain with a continuous wave. Einstein explained them by proposing that light consists of discrete quanta (photons), each with energy $E = hf$, where $f$ is frequency and $h$ is Planck’s constant. An electron is ejected only if a single photon has enough energy to overcome the metal’s work function.

The key concept here is that frequency is an intrinsic property of the wave set by the source, not by the medium. When light passes from one medium to another (or reflects off a boundary), the number of wave crests arriving per second cannot suddenly change — that would require energy to be created or destroyed. Wavelength and velocity, however, depend on the medium’s refractive index, so they can (and do) change upon refraction. Amplitude is affected by the fraction of energy reflected or transmitted, so it also varies.

Let’s walk through each option:

1. Frequency (A) — The source emits light at a fixed frequency. When the wave hits the boundary, the oscillations of the electric and magnetic fields must match on both sides. For reflection, the wave stays in the same medium, so frequency is obviously unchanged. For refraction, the wave enters a new medium, but the boundary condition forces the frequency to remain the same — otherwise, the fields would not be continuous across the interface. So frequency is invariant in both cases.

2. Wavelength (B) — Wavelength is related to frequency and velocity by $\lambda = v / f$. Since $f$ is constant but $v$ changes when light enters a different medium (e.g., from air to glass, speed decreases), the wavelength must also change. In reflection, the wave stays in the same medium, so wavelength is unchanged there — but the question asks for a property that is the same in both reflected and refracted light. Because refraction changes wavelength, this is not the answer.

3. Velocity (C) — The speed of light depends on the medium’s refractive index: $v = c/n$. Reflected light stays in the original medium, so its speed is unchanged. Refracted light enters a new medium, so its speed changes. Thus velocity is not the same for both.

4. Amplitude (D) — Amplitude is related to the energy carried by the wave. At a boundary, some energy is reflected and some transmitted (refracted). The reflected and refracted amplitudes depend on the angle of incidence and the refractive indices (via Fresnel equations). They are generally different from each other and from the incident amplitude. So amplitude is not the same.

The concept first

When a wave meets a boundary, ask three questions in this order.

1. What does the source control? The frequency. The particles at the boundary are forced to oscillate at the frequency of the wave arriving on them; they in turn drive the particles of the second medium at that same rate. So $f$ is a property of the source, and it is invariant across any boundary. This is the anchor of the whole problem.

2. What does the medium control? The speed. For a mechanical wave, $v = \sqrt{\text{elasticity}/\text{inertia}}$; a rarer medium has less inertia per unit volume, so the wave moves faster in it. (For light, rarer = smaller refractive index, and $v = c/n$ is again larger.)

3. What must follow? From $v = f\lambda$, with $f$ locked,

so a faster wave necessarily has a longer wavelength.

Step-by-step

Step 1 — D) Frequency. Unchanged at a boundary. → III (will remain unchanged).

Step 2 — A) Speed. Denser → rarer means the wave speeds up: $v_2 > v_1$. → I (will increase).

Step 3 — B) Wavelength. $\lambda_2/\lambda_1 = v_2/v_1 > 1$. → I (will increase).

Step 4 — C) Amplitude. Use the standard transmission coefficient for a wave crossing into a medium of speed $v_2$:

Because $v_2 > v_1$, the bracket is greater than 1 (it tends to 2 as $v_2 \gg v_1$). So the transmitted amplitude increases. → I (will increase).

(Energy is still conserved — part of it goes into the reflected wave, and the rarer medium carries a given energy with a larger displacement because its impedance is lower.)

Step 5 — Assemble the match. A–I, B–I, C–I, D–III → option (C). Every other option either lets the frequency change or has the speed decreasing, both of which contradict the two anchors above.

ANSWER: C

Why electromagnetic waves have a special structure

An electromagnetic wave is a self-sustaining disturbance in which oscillating electric and magnetic fields regenerate each other as they travel through space. Maxwell's equations impose strict constraints on how these fields must be arranged:

1. Transverse nature: Both $\vec{E}$ and $\vec{B}$ must be perpendicular to the direction of wave propagation.
2. Mutual perpendicularity: $\vec{E}$ and $\vec{B}$ must be perpendicular to each other.
3. Phase relationship: The two fields must oscillate in phase (reach maxima and minima together).
4. Direction relationship: The propagation direction is given by $\vec{E} \times \vec{B}$.

These aren't arbitrary choices but emerge directly from the wave equations derived from Maxwell's equations in free space.

Checking each option systematically

1. Option (A): $\vec{E} = E_0 \sin \omega \left( t - \frac{x}{c} \right) \hat{j}$, $\vec{B} = B_0 \sin \omega \left( t - \frac{x}{c} \right) \hat{k}$

   The argument $\left( t - \frac{x}{c} \right)$ tells us the wave propagates in the $+x$ direction. The electric field oscillates along $\hat{j}$ (the $y$-axis) and the magnetic field along $\hat{k}$ (the $z$-axis). These are perpendicular to each other and both perpendicular to $\hat{i}$ (the propagation direction). The cross product $\hat{j} \times \hat{k} = \hat{i}$ confirms the wave travels in the $+x$ direction. Both fields have identical phase (same sine function), so they oscillate together. ✓

2. Option (B): $\vec{E} = E_0 \sin \omega \left( t - \frac{y}{c} \right) \hat{j}$, $\vec{B} = B_0 \sin \omega \left( t - \frac{z}{c} \right) \hat{k}$

   The electric field depends on $y$ while oscillating along $\hat{j}$, meaning it varies in its own direction of oscillation—this violates the transverse requirement. Similarly, $\vec{B}$ depends on $z$ while pointing along $\hat{k}$. Moreover, the two fields have different arguments, so they propagate in different directions ($+y$ and $+z$ respectively). This cannot represent a single electromagnetic wave. ✗

3. Option (C): $\vec{E} = E_0 \sin \omega \left( t - \frac{x}{c} \right) \hat{i}$, $\vec{B} = B_0 \sin \omega \left( t - \frac{x}{c} \right) \hat{i}$

   Both fields point in the same direction ($\hat{i}$), so they are parallel, not perpendicular. This violates the fundamental requirement that $\vec{E} \perp \vec{B}$ in an electromagnetic wave. Additionally, both fields point along the direction of propagation ($x$-direction), violating the transverse nature. ✗

4. Option (D): $\vec{E} = E_0 \sin^2 \omega \left( t - \frac{x}{c} \right) \hat{j}$, $\vec{B} = B_0 \sin^2 \omega \left( t - \frac{x}{c} \right) \hat{k}$

   The directions and propagation are correct, but the functional form is wrong. Using $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$, we see this represents a wave with a DC offset and double the frequency. More critically, $\sin^2$ is always non-negative, so the fields never reverse direction—they only vary in magnitude. A proper electromagnetic wave requires the fields to oscillate symmetrically about zero. ✗

The most common mistake students make in this problem is to apply Snell’s law or the refractive index formula to frequency. But frequency is a fundamental property of the source — it is set by the oscillation of the electrons in the atom that emitted the light. When light passes from one medium to another, the frequency stays constant because the wave crests cannot pile up or disappear at the boundary. What changes are the speed and the wavelength, since the medium slows the wave down.

So the key idea is: frequency in medium = frequency in vacuum. We just need to calculate the vacuum frequency from the given vacuum wavelength.

1. Find the frequency in vacuum. In vacuum, the wave equation is $c = f \lambda_0$, where $c = 3 \times 10^8$ m/s and $\lambda_0 = 450$ nm $= 450 \times 10^{-9}$ m.

2. Frequency in the medium is unchanged. The refractive index $n = 1.5$ tells us that the speed in the medium is $v = c/n = 2 \times 10^8$ m/s, and the wavelength in the medium becomes $\lambda = \lambda_0 / n = 300$ nm. But the frequency $f$ remains exactly $6.67 \times 10^{14}$ Hz — it does not get divided or multiplied by $n$.