This question has a paragraph followed by two statements, Statement I and Statement II. Of the given

Q: Why is there no phase change when light reflects from the curved surface of the lens?

Because the light is traveling inside the glass (denser medium) and reflecting off the boundary with air (rarer medium). Phase changes of $\pi$ only occur when reflecting off a denser medium.

Q: What would happen if we filled the gap with a liquid denser than glass?

If the liquid is denser than glass, the first reflection (glass to liquid) would have a $\pi$ phase change, and the second reflection (liquid to glass) would have no phase change. The center would still be dark! But if the liquid's refractive index is between air and glass, both reflections would have a $\pi$ phase change, making the center bright.

Q: Why do we use a plano-convex lens of large radius of curvature?

A large radius of curvature makes the air film thickness change very gradually, which spreads out the interference fringes (rings) so they are wide enough to be easily visible under a microscope.

Q: Is Statement II a direct consequence of Statement I?

Yes! The center is dark precisely because of the $\pi$ phase change introduced by the reflection at the air-glass interface (Statement I). Without this phase change, the path difference at the center ($t=0$) would be zero, resulting in a bright spot.

Q: Can we observe Newton's rings with white light?

Yes, but instead of alternating dark and bright rings, you will see colored rings. The center, however, will still be dark because the condition for destructive interference at $t=0$ holds true for all wavelengths.

Question

This question has a paragraph followed by two statements, Statement I and Statement II. Of the given four alternatives after the statements, choose the one that describes the statements.
A thin air film is formed by putting the convex surface of a plane-convex lens over a plane glass plate. With monochromatic light, this film gives an interference pattern due to light reflected from the top (convex) surface and the bottom (glass plate) surface of the film.
Statement I: When light reflects from the air-glass plate interface, the reflected wave suffers a phase change of $\pi$.
Statement II: The centre of the interference pattern is dark.

Options:

(A) Statement I is true, Statement II is true; Statement II is the correct explanation of Statement I

(B) Statement I is true, Statement II is true; Statement II is not the correct explanation of Statement I

(C) Statement I is false, Statement II is true

(D) Statement I is true, Statement II is false

Sigma Percentile · Accepted Answer

The phenomenon of Newton's rings is one of the most beautiful demonstrations of wave optics, revealing the hidden wave nature of light through a simple arrangement of glass. Let's dive deep into the physics of this setup and unravel the mystery of its dark center.

### Analyzing the Setup

Imagine a plano-convex lens resting gently on a flat glass plate. Because the lens is curved, it doesn't sit perfectly flush against the flat plate. Instead, it touches the plate at exactly one point—the center—and curves away from it everywhere else.

This geometry creates a very thin, wedge-shaped film of air trapped between the bottom of the lens and the top of the glass plate. When monochromatic light shines down from above, it encounters this air film and undergoes multiple reflections, setting the stage for interference.

### The Physics of Reflection

To understand the interference pattern, we must trace the path of the light rays. When a ray of light travels down through the lens, it hits the first boundary: the interface between the bottom of the glass lens and the air film. Here, some light reflects back up. Because the light is traveling from a **denser medium (glass)** and reflecting off a **rarer medium (air)**, this reflection occurs with **no phase change**.

However, the portion of light that transmits through the air film eventually hits the second boundary: the interface between the air film and the flat glass plate below. When this light reflects back up, it is reflecting from a **rarer medium (air)** off a **denser medium (glass)**. According to Stokes' principle in wave optics, a reflection at a rarer-to-denser boundary introduces a sudden **phase change of $\pi$** (which is equivalent to a path difference of $\frac{\lambda}{2}$).

This confirms that **Statement I is absolutely true**.

### The Master Equation

The two reflected rays—one from the top of the air film and one from the bottom—now travel back up and interfere with each other. The nature of their interference depends on their total effective path difference, $\Delta x$.

The path difference is determined by the extra distance the second ray traveled through the air film (which is $2t$, where $t$ is the thickness of the film) plus the effective path difference introduced by the phase change.

$$ \Delta x = 2\mu t + \frac{\lambda}{2} $$

Since the film is made of air, the refractive index $\mu \approx 1$, simplifying our equation to:

$$ \Delta x = 2t + \frac{\lambda}{2} $$

### The Dark Center

Now, let's focus our attention on the exact center of the setup, where the convex lens physically touches the flat glass plate. At this precise point of contact, the thickness of the air film is exactly zero ($t = 0$).

Let's substitute this into our master equation:

$$ \Delta x = 2(0) + \frac{\lambda}{2} = \frac{\lambda}{2} $$

A path difference of exactly half a wavelength ($\frac{\lambda}{2}$) means that the two reflected waves are perfectly out of phase. When the peak of one wave aligns with the trough of the other, they cancel each other out completely. This is the condition for **destructive interference**.

As a result, the central spot of the interference pattern will be completely dark. This confirms that **Statement II is also true**.

### The Logical Connection

We have established that both statements are factually correct. But how do they relate to each other?

Does the fact that the center is dark explain why there is a phase change? Absolutely not. That is backward logic. The phase change of $\pi$ upon reflection is a fundamental property of electromagnetic waves interacting with boundaries. It is this very phase change that *causes* the path difference to be $\frac{\lambda}{2}$ at the center, which in turn *causes* the center to be dark.

Therefore, **Statement I explains Statement II**. This means that Statement II is **not** the correct explanation for Statement I.

Both statements are true, but Statement II does not explain Statement I.

Select Answer:

Visualized Solution

Newton's Rings Setup

Reflection and Phase Change

Path Difference at the Center

Central Fringe Condition

Conclusion

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

Analyzing the Setup

The Physics of Reflection

The Master Equation

The Dark Center

The Logical Connection

Similar Questions

In a double slit experiment, when a thin film of thickness $t$ having refractive index $μ$ is introduced in front of one of the slits, the maximum at the centre of the fringe pattern shifts by one fringe width. The value of $t$ is ( $λ$ is the wavelength of the light used)

List-I

List-II