Statement I: On viewing the clear, blue portion of the sky through a Calcite crystal, the intensity

Q: Why does blue light scatter the most?

According to Rayleigh scattering, the intensity of scattered light is inversely proportional to the fourth power of the wavelength ($I \propto 1/\lambda^4$). Since blue light has a shorter wavelength than red light, it scatters much more strongly.

Q: Is all light from the sky completely polarized?

No, only the light scattered at exactly $90^\circ$ to the incident sunlight is completely plane-polarized. Light scattered at other angles is only partially polarized.

Q: What is a Calcite crystal?

Calcite is a naturally occurring birefringent crystal. It has the special property of splitting unpolarized light into two polarized rays, effectively acting as a polarizer.

Q: What if the light from the sky was unpolarized?

If the light entering the Calcite crystal were completely unpolarized, the transmitted intensity would be a constant $I_0/2$, regardless of how you rotate the crystal. The variation in intensity proves the incoming light is polarized!

Q: Why does the intensity go to zero at certain angles?

According to Malus's Law, $I = I_0 \cos^2\theta$. When the pass axis of the crystal is perpendicular to the polarization direction of the light ($\theta = 90^\circ$ or $270^\circ$), $\cos\theta = 0$, so no light is transmitted.

Question

Statement I: On viewing the clear, blue portion of the sky through a Calcite crystal, the intensity of transmitted light varies as the crystal is rotated.
Statement II: The light coming from the sky is polarised due to scattering of sun light by particles in the atmosphere. The scattering is largest for blue light.

Options:

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

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

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

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

Sigma Percentile · Accepted Answer

### The Hidden Physics of the Blue Sky

Imagine looking up at the clear blue sky on a sunny day. Have you ever wondered why it is blue, or what invisible properties that light carries before it reaches your eyes? The answer lies in a beautiful interplay of scattering and polarization.

Sunlight, as it travels through the vacuum of space, is an electromagnetic wave. It is **unpolarized**, meaning its electric field vectors oscillate randomly in all possible directions perpendicular to its path. However, the moment this light enters Earth's atmosphere, everything changes.

### Rayleigh Scattering and Polarization

Our atmosphere is filled with tiny gas molecules and particles. When sunlight hits these particles, it scatters in all directions. According to **Rayleigh Scattering**, the intensity of scattered light is inversely proportional to the fourth power of its wavelength ($I \propto \frac{1}{\lambda^4}$).

Because blue light has a much shorter wavelength than red light, it scatters significantly more, painting our sky blue. But there is a hidden secret here: light that is scattered at exactly $90^\circ$ to the incident sunlight becomes **completely plane-polarized**! This means its electric field vectors are now oscillating in only one single plane.

### The Calcite Crystal Experiment

Now, let's bring in the Calcite crystal mentioned in Statement I. Calcite ($CaCO_3$) is a naturally occurring birefringent material. In optics, it acts as a **polarizer**. A polarizer is like a picket fence for light; it only allows light waves oscillating parallel to its "pass axis" to transmit through.

If we look at the $90^\circ$ scattered blue light from the sky through this Calcite crystal, we are essentially passing already polarized light through a polarizer.

### Malus's Law in Action

What happens when polarized light passes through a polarizer? The transmitted intensity is governed by **Malus's Law**:

$$I = I_0 \cos^2	heta$$

Here, $I_0$ is the initial intensity of the polarized light, and $	heta$ is the angle between the light's polarization direction and the pass axis of the Calcite crystal.

If we start rotating the Calcite crystal while looking through it, the angle $	heta$ changes continuously. Because the $\cos^2	heta$ term oscillates between $0$ and $1$, the transmitted intensity $I$ will also vary periodically. It will reach a maximum of $I_0$ when the axes align ($	heta = 0, \pi$), and it will drop to absolute zero when they are perpendicular ($	heta = \pi/2, 3\pi/2$).

### Conclusion

Let's evaluate the statements based on our physical journey.

**Statement I** claims that the intensity of transmitted light varies as the crystal is rotated. As we proved using Malus's Law, this is absolutely **True**.

**Statement II** claims that the light from the sky is polarized due to scattering, which is largest for blue light. As we saw with Rayleigh scattering, this is also **True**.

Most importantly, the only reason the Calcite crystal was able to vary the intensity of the light upon rotation is *because* the incoming light was already polarized by the atmosphere. Therefore, **Statement II is the correct explanation for Statement I**.

Select Answer:

Visualized Solution

$U n p o l a r i z e d S u n l i g h t$

$R a y l e i g h S c a tt er in g & P o l a r i z a t i o n$

$C a l c i t e C r y s t a l a s P o l a r i z er$

$M a l u s^{'} s L a w$

$R o t a t in g t h e C r y s t a l$

$I n t e n s i t y V a r ia t i o n$

$C o n c l u s i o n$

The Sigma Insight: Polarization

The Hidden Physics of the Blue Sky

Rayleigh Scattering and Polarization

The Calcite Crystal Experiment

Malus's Law in Action

Conclusion

Similar Questions

In the List-I, four optical effects are mentioned. The physical phenomena of light which are essential to describe these optical effects are given in List-II. Choose the option which describes the correct match between the entries in List-I to those in List-II.

List-I

List-II

When an unpolarised light of intensity $I_{0}$ is incident on a polarising sheet, the intensity of the light which does not get transmitted is

The angle of incidence at which reflected light is totally polarised for reflection from air to glass (refractive index $n$ ), is

A beam of unpolarised light of intensity $I_{0}$ is passed through a polaroid $A$ and then through another polaroid $B$ which is oriented so that its principal plane makes an angle of $4 5^{\circ}$ relative to that of $A$ . The intensity of the emergent light is