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: Frequency Invariance — when light passes from one medium to another, its frequency remains unchanged because it is determined by the source. Only speed and wavelength change.

Step 1: Reflected light

Reflection occurs in the same medium (air). So speed $c = 3 \times 10^8\ \text{m/s}$, wavelength $\lambda = 589\ \text{nm}$, and frequency

Step 2: Refracted light

In water, speed reduces:

Frequency stays the same: $f = 5.09 \times 10^{14}\ \text{Hz}$.

Wavelength in water, computed directly as $\lambda' = \lambda_0/n$ (the exact relation — dividing the already-rounded $v$ by the already-rounded $f$ instead would introduce a small rounding error):

The single most important idea in this problem is frequency invariance. When light crosses from one medium to another, its frequency does not change. Why? Because frequency is set by the source — the oscillating charges in the light source emit a certain number of wave crests per second. When that wave enters a different medium, the crests cannot pile up or vanish at the boundary; they must arrive and depart at the same rate. So the frequency stays the same in air, water, or any transparent medium.

What does change is the speed of light, and consequently the wavelength. In a medium of refractive index $n$, light travels slower: $v = c/n$. Since $v = f\lambda$, if $f$ is fixed and $v$ drops, $\lambda$ must also drop by the same factor.

Let's apply this cleanly.

Given data:

• Wavelength in air (vacuum essentially): $\lambda_0 = 589\ \text{nm} = 589 \times 10^{-9}\ \text{m}$
• Speed of light in vacuum/air: $c = 3 \times 10^8\ \text{m/s}$
• Refractive index of water: $n = 1.33$

1. Find the frequency in air (which will be the same everywhere)

Frequency is the only quantity we can compute directly from the air values:

Do the division:

This frequency is the same for both reflected and refracted light.

2. (a) Reflected light

Reflection occurs at the air-water boundary, but the reflected ray stays in air. So the medium of propagation is unchanged.

• Speed: $v_{\text{reflected}} = c = 3 \times 10^8\ \text{m/s}$
• Frequency: $f = 5.09 \times 10^{14}\ \text{Hz}$ (invariant)
• Wavelength: $\lambda_{\text{reflected}} = \lambda_0 = 589\ \text{nm}$

No calculation needed — reflected light is still in air, so all wave parameters are identical to the incident wave.

3. (b) Refracted light

The refracted ray enters water. Now the speed changes:

Compute:

Frequency remains $f = 5.09 \times 10^{14}\ \text{Hz}$.

The cleanest way to get the wavelength in water is directly from $\lambda = \lambda_0/n$ (since $v = c/n$ and $f$ is constant, $\lambda = v/f = (c/n)/f = \lambda_0/n$ exactly):

4. Summary table

Method: Law of Reflection + Snell’s Law + Frequency Invariance Principle

Core Concept (Why this works)

When light crosses a boundary, frequency never changes — it is determined by the source, not the medium. Only wavelength and speed change according to the refractive index.

Step 1 — Identify given data

• Wavelength in air: $\lambda_a = 589\ \text{nm} = 589 \times 10^{-9}\ \text{m}$
• Speed of light in air (vacuum): $c = 3 \times 10^8\ \text{m/s}$
• Refractive index of water: $n_w = 1.33$
• Refractive index of air: $n_a \approx 1$

Step 2 — Find frequency (same for both reflected and refracted)

Using the relation in air:

This frequency remains identical for reflected and refracted light.

Step 3 — (a) Reflected light

• Reflection occurs in the same medium (air).
• Speed: $v_{\text{reflected}} = c = \boxed{3 \times 10^8\ \text{m/s}}$
• Wavelength: $\lambda_{\text{reflected}} = \lambda_a = \boxed{589\ \text{nm}}$
• Frequency: $\boxed{5.09 \times 10^{14}\ \text{Hz}}$ (as above)

Step 4 — (b) Refracted light (enters water)

• Speed changes:

• Wavelength changes proportionally:

• Frequency remains: $\boxed{5.09 \times 10^{14}\ \text{Hz}}$

Final Answer Summary

Key takeaway: Frequency is invariant across boundaries — always calculate it first from the source medium, then use $v = \nu \lambda$ and $n = c/v$ to find the other quantities in each medium.

Here are the common mistakes students make on this exact problem, along with the concept-first reasoning to avoid each.

Mistake 1: Thinking frequency changes when light enters water

The error:

Students often assume that because speed and wavelength change in a medium, frequency must also change. They then try to calculate a new frequency using $f = v/\lambda$ with the new speed and wavelength — getting a wrong answer.

Why it’s wrong:

Frequency is determined by the source (the original light wave), not the medium. When light crosses a boundary, the wave crests arrive at the same rate they left — frequency is invariant.

How to avoid:

• Memorise the rule: Frequency never changes on reflection or refraction.
• Always calculate frequency from the given vacuum/air data first:

• Then carry this same $f$ into both reflected and refracted parts.

Mistake 2: Forgetting that reflection keeps the same medium

The error:

Some students treat reflected light as if it enters water — they change its speed and wavelength.

Why it’s wrong:

Reflection occurs at the air-water boundary, but the reflected ray stays in air. So its speed and wavelength remain identical to the incident light.

How to avoid:

• Draw a clear ray diagram. Label the reflected ray as staying in air.
• For part (a), simply state:
  • Speed = $c = 3 \times 10^8\ \text{m/s}$
  • Wavelength = $589\ \text{nm}$
  • Frequency = same as above ($5.09 \times 10^{14}\ \text{Hz}$)

Mistake 3: Using the wrong formula for wavelength in water

The error:

Students sometimes write $\lambda_{\text{water}} = \frac{\lambda_{\text{air}}}{n}$ but forget to check whether $n$ is the refractive index of water relative to air (it is, here). Others mistakenly use $n = \frac{\lambda_{\text{water}}}{\lambda_{\text{air}}}$ (inverted).

Why it’s correct:

Refractive index is defined as:

So:

How to avoid:

• Always write the definition: $n = \frac{\text{speed in vacuum}}{\text{speed in medium}} = \frac{\lambda_{\text{vacuum}}}{\lambda_{\text{medium}}}$
• Then solve for the unknown. If $n > 1$, wavelength decreases — that’s a quick sanity check.

Mistake 4: Calculating speed in water incorrectly

The error:

Using $v = c \times n$ instead of $v = c / n$, or forgetting that $n=1.33$ means light slows down.

Why it’s wrong:

$n$ is always $\geq 1$ for ordinary media. Speed in medium is less than $c$:

How to avoid:

• Remember: $n$ is a ratio of speeds — light is slower in denser media.
• Use dimensional check: $c$ has units m/s, dividing by a pure number gives m/s — correct.

Mistake 5: Mixing up reflected vs. refracted quantities in the final answer

The error:

Writing the reflected light’s wavelength as $443\ \text{nm}$ or the refracted light’s speed as $3 \times 10^8\ \text{m/s}$.

How to avoid:

• Make two clear columns or bullet lists in your answer:
  • (a) Reflected (in air): $v = c$, $\lambda = 589\ \text{nm}$, $f = 5.09 \times 10^{14}\ \text{Hz}$
  • (b) Refracted (in water): $v = c/n$, $\lambda = \lambda_{\text{air}}/n$, $f = \text{same as incident}$

Quick Summary Table (Exam-Ready)

Final tip: Always start any such problem by calculating frequency from the given vacuum/air data — that single number is the key to both parts.