A point source S emits unpolarized light uniformly in all directions. At two points A and B, the rat

Q: Why does the first polaroid exactly halve the intensity of the unpolarized light?

Unpolarized light has electric field vectors oscillating randomly in all directions perpendicular to propagation. When you resolve these random vectors into components parallel and perpendicular to the polaroid's pass axis, the average of $\cos^2 \theta$ over all possible angles from $0$ to $2\pi$ is exactly $1/2$. Therefore, half the energy passes through, and half is absorbed.

Q: Does the distance of the polaroids from the source matter for this calculation?

No, it doesn't. The problem states they are placed 'just before point B'. This means the intensity of the light right as it hits the first polaroid is exactly $I_B$. We only care about how the polaroids modify this specific intensity.

Q: Why didn't we use the inverse square law ($I \propto 1/r^2$) to find the exact distances?

The inverse square law is the underlying physical reason why $I_A$ and $I_B$ are different initially. However, since the problem directly gives us their initial ratio ($I_A/I_B = 2$), we can bypass calculating the exact distances $r_A$ and $r_B$. We only need to track how $I_B$ changes relative to its initial state.

Q: What if the angle between the two polaroids was $90^\circ$ instead of $45^\circ$?

If the polaroids were 'crossed' at $90^\circ$, then $\cos 90^\circ = 0$. According to Malus's Law, no light would pass through the second polaroid. The final intensity at B ($I_B'$) would be zero, making the new ratio $r'$ approach infinity.

Q: Is the intensity at point A affected by placing polaroids near point B?

Not at all. The polaroids are placed exclusively in the path of the light traveling towards point B. The light rays traveling towards point A remain completely unobstructed, so $I_A$ stays exactly the same throughout the problem.

Question

A point source S emits unpolarized light uniformly in all directions. At two points A and B, the ratio $r = I_A/I_B$ of the intensities of light is 2. If a set of two polaroids having $45^\circ$ angle between their pass-axes is placed just before point B, then the new value of r will be _____.

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Points

Imagine a tiny, brilliant bulb—a point source $S$—radiating unpolarized light equally in all directions into the vastness of space. As the light waves travel outward, they spread over larger and larger spherical areas, causing their intensity to drop according to the inverse square law ($I \propto \frac{1}{r^2}$).

We are observing this light at two distinct locations: point $A$ and point $B$. The problem provides us with a crucial piece of initial intelligence: the intensity at point $A$ is exactly twice the intensity at point $B$. Mathematically, we can write this initial ratio as:

$$r = \frac{I_A}{I_B} = 2$$

This ratio is our anchor. It tells us the relative brightness of the two points before we start messing with the light's path.

### The First Gatekeeper: Unpolarized to Polarized

Now, we introduce a twist. Just before the light reaches point $B$, we place a set of two polaroids in its path. Think of polaroids as optical gatekeepers. They only allow light waves oscillating in a specific direction (their pass-axis) to go through.

The light arriving from our source $S$ is **unpolarized**, meaning its electric field vectors are oscillating randomly in all possible directions perpendicular to the direction of travel. When this chaotic, unpolarized light of intensity $I_B$ hits the first polaroid ($P_1$), a remarkable filtering happens.

No matter how $P_1$ is oriented, it will block exactly half of the unpolarized light. The light that emerges is now perfectly polarized along the pass-axis of $P_1$. The intensity of this newly polarized light, let's call it $I_1$, is simply half of what arrived:

$$I_1 = \frac{I_B}{2}$$

### The Second Gatekeeper: Malus's Law in Action

But the journey isn't over. This polarized light immediately encounters the second polaroid ($P_2$). The problem states that the pass-axis of $P_2$ is tilted at an angle of $	heta = 45^\circ$ relative to $P_1$.

To find out how much light survives this second gatekeeper, we must invoke **Malus's Law**. This fundamental law of optics states that when completely polarized light is incident on an analyzer (our second polaroid), the transmitted intensity is proportional to the square of the cosine of the angle between their transmission axes.

Let's set up the master equation for the new intensity at point $B$, which we will call $I_B'$:

$$I_B' = I_1 \cos^2 	heta$$

Substituting our known values into this raw structure:

$$I_B' = \left(\frac{I_B}{2}ight) \cos^2 45^\circ$$

### The Final Reckoning: Calculating the New Ratio

Now, it's time for the atomic computation. We know from basic trigonometry that $\cos 45^\circ = \frac{1}{\sqrt{2}}$. Squaring this gives us $\frac{1}{2}$. Let's plug that in:

$$I_B' = \frac{I_B}{2} 	imes \left(\frac{1}{\sqrt{2}}ight)^2$$

$$I_B' = \frac{I_B}{2} 	imes \frac{1}{2} = \frac{I_B}{4}$$

The dual polaroid setup has effectively reduced the intensity of light reaching point $B$ to a quarter of its original value!

Finally, we need to find the new ratio, $r'$, which is the intensity at $A$ divided by the new intensity at $B$. Remember, the light reaching point $A$ was completely unobstructed, so $I_A$ remains unchanged.

$$r' = \frac{I_A}{I_B'}$$

Substitute our expression for $I_B'$:

$$r' = \frac{I_A}{\frac{I_B}{4}}$$

The $4$ in the denominator flips up to the numerator:

$$r' = 4 \left(\frac{I_A}{I_B}ight)$$

Since our initial anchor ratio $\frac{I_A}{I_B}$ was $2$, we simply multiply:

$$r' = 4 	imes 2 = 8$$

The new ratio of the intensities is **8**. By understanding the sequential filtering of light through polaroids, we've elegantly arrived at the final answer.

Enter Numerical Value:

Visualized Solution

Initial Setup

Inserting Polaroids

Intensity after $P_{1}$

Malus's Law for $P_{2}$

Final Intensity at $B$

New Ratio $r^{'}$

What if Source was Polarized?

The Sigma Insight: Polarization

The Setup

The First Gatekeeper

The Second Gatekeeper

The Final Reckoning

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