Animated Solution for Physics - Optics: Unpolarised light of intensity $I$ passes through an ideal polariser $A$. Another identical polariser $B$ is placed behind $A$. The intensity of light beyond $B$ is found to be $\frac{I}{2}$. Now, another identical polariser $C$ is placed between $A$ and $B$. The intensity beyond $B$ is now found to be $\frac{I}{8}$. The angle between polariser $A$ and $C$ is

Unpolarised to Polarised

Let's visualize the initial setup of our optical system. We begin with a beam of unpolarised light possessing an initial intensity of $I$. This light approaches our first ideal polariser, which we'll call polariser $A$.

When unpolarised light passes through an ideal polariser, a fundamental rule of optics dictates that exactly half of its intensity is transmitted. The polariser acts as a filter, only allowing the component of the electric field parallel to its transmission axis to pass through. Therefore, the intensity right after polariser $A$ becomes:

Next, the problem introduces a second identical polariser, $B$, placed directly behind $A$. We are given a crucial piece of information: the intensity of the light beyond $B$ is still $\frac{I}{2}$.

What does this tell us about the physical orientation of the polarisers? Since the intensity did not drop at all after passing through $B$, it means that $100\%$ of the light that made it through $A$ also made it through $B$. According to Malus's Law, this is only possible if the transmission axes of polarisers $A$ and $B$ are perfectly parallel to each other. Thus, the angle between them is $0^{\circ }$.

Polariser C

Now comes the twist in the experiment. We insert a third polariser, $C$, right between $A$ and $B$. Let's assume the transmission axis of $C$ is rotated by an unknown angle $\alpha$ relative to polariser $A$.

According to Malus's Law, the intensity of light transmitted through polariser $C$ depends heavily on this angle $\alpha$. The intensity after $C$ will be the incoming intensity from $A$ multiplied by the cosine squared of the angle between their axes:

This newly polarised light from $C$ now travels towards polariser $B$. Remember our earlier deduction: $A$ and $B$ are perfectly parallel. Therefore, if $C$ is at an angle $\alpha$ with $A$, it must also be at the exact same angle $\alpha$ with $B$.

Applying Malus's Law one more time for the light passing from $C$ to $B$, the final intensity after $B$ will be the intensity after $C$ multiplied by the cosine squared of $\alpha$:

Let's substitute the value of $I_C$ we found earlier into our final equation. We get the final intensity as:

This expression simplifies beautifully to:

The problem gives us the final piece of the puzzle: the new intensity beyond $B$ is now $\frac{I}{8}$. Equating our theoretical expression to this given value, we can set up our master equation:

We can immediately cancel out the initial intensity $I$ from both sides. Multiplying both sides by $2$, we find that:

We are almost at the finish line. Taking the square root of both sides gives:

Taking the square root one more time yields:

From our knowledge of trigonometry, we know that the angle whose cosine is $\frac{1}{\sqrt{2}}$ is exactly $45^{\circ }$. Therefore, the angle between polariser $A$ and $C$ is $45^{\circ }$.