Two light waves having the same wavelength lambda in vacuum are in phase initially. Then, the first

Q: Why does the wavelength change in a medium but not the frequency?

Frequency is a property of the source. Think of it like a drummer beating a drum; the beats per second don't change. But since light slows down in a medium ($v = c/n$), the distance between consecutive crests (wavelength) must compress to maintain the same beat rate ($v = f\lambda$).

Q: What exactly is 'Optical Path Length'?

It's the equivalent distance light would have traveled in a perfect vacuum in the same amount of time it took to travel through the medium. Mathematically, it's $n \times L$. It allows us to compare paths through different materials on a level playing field.

Q: Why do we subtract $\phi_2$ from $\phi_1$ and not the other way around?

Phase difference is usually taken as a magnitude, or it depends on which wave you consider as the reference. The options provided in the question dictate the format. Both $\phi_1 - \phi_2$ and $\phi_2 - \phi_1$ represent the same physical reality, just with a different sign convention.

Q: Does the physical distance $L$ have to be the same for both waves?

Not at all! The beauty of this formula is that $L_1$ and $L_2$ can be completely different. The phase difference depends on the *optical* path difference, which accounts for both the physical distance and the refractive index.

Q: What if the waves didn't start in phase?

If they had an initial phase difference, say $\phi_0$, you would simply add that initial difference to the final calculated difference. The journey through the mediums just *adds* to whatever phase relationship they already had.

Question

Two light waves having the same wavelength $\lambda$ in vacuum are in phase initially. Then, the first wave travels a path $L_1$ through a medium of refractive index $n_1$ while the second wave travels a path of length $L_2$ through a medium of refractive index $n_2$. After this, the phase difference between the two waves is

Options:

(A) $\frac{2\pi}{\lambda}(n_1 L_1 - n_2 L_2)$

(B) $\frac{2\pi}{\lambda}\left(\frac{L_1}{n_1} - \frac{L_2}{n_2}\right)$

(C) $\frac{2\pi}{\lambda}\left(\frac{L_2}{n_1} - \frac{L_1}{n_2}\right)$

(D) $\frac{2\pi}{\lambda}(n_2 L_1 - n_1 L_2)$

Sigma Percentile · Accepted Answer

### The Dance of the Waves

Imagine two identical light waves, born at the exact same moment in the vast emptiness of a perfect vacuum. They are perfectly in sync, their crests and troughs rising and falling together in a beautiful, synchronized dance. In physics, we say these waves are **in phase**.

But what happens when their paths diverge? Suppose the first wave plunges into a block of glass (a medium with refractive index $n_1$) and travels a distance $L_1$. Meanwhile, the second wave dives into a pool of water (a medium with refractive index $n_2$) and travels a distance $L_2$.

Because light travels at different speeds in different materials, our two waves will no longer be perfectly synchronized when they emerge. One will have fallen behind the other. The question we must answer is: exactly how out of sync are they? What is their new **phase difference**?

### The Mathematics of Phase Accumulation

To solve this mystery, we need to understand how a wave accumulates phase as it travels. Think of phase as a clock that ticks forward as the wave moves through space. For every full wavelength $\lambda$ the wave completes, the phase clock ticks forward by exactly $2\pi$ radians.

Therefore, if a wave travels a total distance $L$, the number of full wavelengths it completes is simply $\frac{L}{\lambda}$. To find the total phase $\phi$ accumulated over this distance, we multiply the number of waves by $2\pi$:

$$ \phi = \frac{2\pi}{\lambda} L $$

This formula works perfectly in a vacuum. But our waves are traveling through mediums!

### The Wavelength Compression Effect

Here is the critical catch: when light enters a medium with a refractive index $n$, it slows down. However, the frequency of the light (the number of waves passing a point per second) must remain constant. To maintain this constant frequency while traveling slower, the wave must compress.

The new wavelength in the medium, which we will call $\lambda_{med}$, is given by the vacuum wavelength divided by the refractive index:

$$ \lambda_{med} = \frac{\lambda}{n} $$

This means that inside a denser medium, the waves are packed closer together. Because they are packed closer together, more waves will fit into the same physical distance $L$, which means the wave will accumulate phase much faster!

### Calculating the Individual Phases

Now we have all the tools we need. Let's calculate the phase accumulated by the first wave, $\phi_1$, as it travels through its medium. We use our phase formula, but we must be careful to use the compressed wavelength $\lambda_1 = \frac{\lambda}{n_1}$:

$$ \phi_1 = \frac{2\pi}{\lambda_1} L_1 $$

Substituting our expression for $\lambda_1$:

$$ \phi_1 = \frac{2\pi}{\left(\frac{\lambda}{n_1}\right)} L_1 $$

The $n_1$ flips up to the numerator, giving us:

$$ \phi_1 = \frac{2\pi n_1 L_1}{\lambda} $$

We can apply the exact same logic to the second wave. It travels a distance $L_2$ through a medium with refractive index $n_2$. Its accumulated phase $\phi_2$ will be:

$$ \phi_2 = \frac{2\pi n_2 L_2}{\lambda} $$

### The Grand Finale: Phase Difference

We know exactly how much phase each wave has accumulated. To find out how out of sync they are, we simply need to find the difference between their phases, denoted as $\Delta \phi$:

$$ \Delta \phi = \phi_1 - \phi_2 $$

Let's substitute the expressions we just derived:

$$ \Delta \phi = \frac{2\pi n_1 L_1}{\lambda} - \frac{2\pi n_2 L_2}{\lambda} $$

Notice that both terms share a common factor of $\frac{2\pi}{\lambda}$. Let's factor that out to clean up our final equation:

$$ \Delta \phi = \frac{2\pi}{\lambda} (n_1 L_1 - n_2 L_2) $$

And there we have it! This elegant equation tells us exactly how the phase difference depends on the vacuum wavelength, the refractive indices, and the physical distances traveled. Looking at our options, this matches perfectly with option (a).

### The Pro-Tip: Optical Path Length

While our derivation was rigorous and correct, experienced physicists often use a powerful shortcut called **Optical Path Length**.

The optical path length is defined as the physical distance $L$ multiplied by the refractive index $n$.

$$ x_{opt} = nL $$

Conceptually, the optical path length is the distance the light *would have traveled* in a perfect vacuum in the exact same amount of time it took to travel through the medium. It normalizes everything to a vacuum!

Using this concept, the optical path of the first wave is $n_1 L_1$, and the optical path of the second wave is $n_2 L_2$. The **optical path difference** $\Delta x$ is simply:

$$ \Delta x = n_1 L_1 - n_2 L_2 $$

Once you have the optical path difference, you can treat the waves as if they had been traveling in a vacuum the whole time. The phase difference is just $\frac{2\pi}{\lambda}$ times the optical path difference:

$$ \Delta \phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} (n_1 L_1 - n_2 L_2) $$

This shortcut gets you to the answer in seconds and is an incredibly valuable tool to have in your physics arsenal!

Select Answer:

Visualized Solution

Visualizing the Setup

Phase Accumulation Formula

Wavelength in a Medium

Phase of First Wave

Phase of Second Wave

Calculating Phase Difference

Final Answer

The Way Forward: Optical Path

The Sigma Insight: Interference and Young's Double-Slit Experiment

The Dance of the Waves

The Mathematics of Phase Accumulation

The Wavelength Compression Effect

Calculating the Individual Phases

The Grand Finale

The Pro-Tip

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