A beam of plane polarised light of large cross-sectional area and uniform intensity of 3.3 Wm^-2 fal

Q: Why do we take the average of $\cos^2\theta$?

Because the polariser is continuously rotating, the angle $\theta$ changes constantly. The intensity fluctuates between $I_0$ and $0$. To find the total energy over a full cycle, we must use the average intensity during that time.

Q: Why is the average of $\cos^2\theta$ equal to $1/2$?

Mathematically, the integral of $\cos^2\theta$ over a full cycle ($0$ to $2\pi$) divided by $2\pi$ is $1/2$. Physically, $\cos^2\theta + \sin^2\theta = 1$, and over a full cycle, the average of $\cos^2\theta$ and $\sin^2\theta$ must be equal by symmetry, so each must be $1/2$.

Q: Can we integrate instead of taking the average?

Absolutely! You can integrate $dE = I_0 \cos^2(\omega t) \cdot A \cdot dt$ from $t=0$ to $t=T$. Taking the average is just a faster, more intuitive shortcut that yields the exact same result.

Q: Why did we use $31.4$ as $10\pi$?

In physics problems, numbers are often chosen to simplify calculations. Since $\pi \approx 3.14$, $10\pi \approx 31.4$. Recognizing this makes finding the time period $T = 2\pi / \omega$ much easier without a calculator.

Q: What if the incident light was unpolarised?

If the incident light was unpolarised, the first polariser would transmit half the intensity ($I_0/2$) regardless of its rotation angle. The transmitted intensity would be constant, and you wouldn't need to average $\cos^2\theta$ again.

Question

A beam of plane polarised light of large cross-sectional area and uniform intensity of $3.3 	ext{ Wm}^{-2}$ falls normally on a polariser, which rotates about its axis with an angular speed of $31.4 	ext{ rad/s}$. The energy of light passing through the polariser per revolution is close to (Take, cross-sectional area $= 3 	imes 10^{-4} 	ext{ m}^2$)

Options:

(A) $1.5 	imes 10^{-4} 	ext{ J}$

(B) $5.0 	imes 10^{-4} 	ext{ J}$

(C) $1.0 	imes 10^{-4} 	ext{ J}$

(D) $1.0 	imes 10^{-5} 	ext{ J}$

Sigma Percentile · Accepted Answer

The problem of a spinning polariser is a beautiful intersection of optics and kinematics. It challenges us to think beyond static setups and analyze how energy flows when the system itself is in motion. Let's break down this fascinating phenomenon step by step. ### The Spinning Polariser Imagine a beam of plane-polarised light traveling through space. Its electric field vectors are oscillating perfectly in one plane. Suddenly, it encounters a polariser. But this isn't just any polariser—it's spinning about its axis with a constant angular speed $\omega = 31.4 ext{ rad/s}$. Our goal is to find the total energy that manages to pass through this rotating gateway during exactly one complete revolution. To do this, we need to understand how the transmitted light's intensity changes over time. ### Malus's Law in Motion When plane-polarised light hits a polariser, the amount of light that gets through depends on the angle between the light's polarisation plane and the polariser's transmission axis. This relationship is governed by **Malus's Law**, which states that the transmitted intensity $I$ is given by: $$I = I_0 \cos^2 heta$$ Here, $I_0$ is the initial intensity, and $ heta$ is the angle between the two axes. Because our polariser is spinning, this angle $ heta$ is not constant; it's continuously changing as $ heta = \omega t$. This means the transmitted intensity is fluctuating, reaching a maximum when the axes align and dropping to zero when they are perpendicular. ### The Power of Averages Since the intensity is constantly changing, how do we calculate the total energy over a full revolution? We could integrate the intensity over time, but there's a much more elegant shortcut: using the **average intensity**. Over one complete cycle, the average value of the $\cos^2 heta$ function is exactly $\frac{1}{2}$. Therefore, the average intensity $$ passing through the polariser is simply half of the maximum possible intensity: $$ = \frac{I_0}{2}$$ Now, we need to connect this average intensity to energy. Intensity is defined as power per unit area, and power is energy per unit time. Therefore, the total energy $E$ transmitted is the product of the average intensity, the cross-sectional area $A$, and the time $T$ taken for one revolution: $$E = imes A imes T$$ ### Bringing It All Together Before we can plug in our numbers, we need to find the time period $T$ of one revolution. We know that $T = \frac{2\pi}{\omega}$. The problem gives us $\omega = 31.4 ext{ rad/s}$. Notice a clever mathematical trick here: $31.4$ is approximately $10\pi$. Using this approximation makes our calculation beautifully simple: $$T = \frac{2\pi}{10\pi} = \frac{1}{5} ext{ s}$$ Now, we have all our puzzle pieces. Let's substitute them into our energy equation: $$E = \left(\frac{3.3}{2} ight) imes (3 imes 10^{-4}) imes \left(\frac{1}{5} ight)$$ Let's crunch the numbers. Multiplying the numerators, $3.3 imes 3 = 9.9$. Multiplying the denominators, $2 imes 5 = 10$. This gives us: $$E = \frac{9.9}{10} imes 10^{-4} ext{ J}$$ $$E = 0.99 imes 10^{-4} ext{ J}$$ Rounding to the nearest option, we get our final answer: $$E \approx 1.0 imes 10^{-4} ext{ J}$$ And there we have it! By combining Malus's Law with the concept of average values, we've successfully navigated the dynamic world of a spinning polariser.

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Visualized Solution

$I_{0} = 3.3 Wm^{- 2}, ω = 31.4 rad/s$

$Malus’s Law$

$Average Intensity$

$Energy Transmitted$

$Time Period T$

$Substitution$

$Final Calculation$

The Sigma Insight: Polarization

The Spinning Polariser

Malus's Law in Motion

The Power of Averages

Bringing It All Together

Similar Questions

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

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

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I0​=3.3 Wm−2,ω=31.4 rad/s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Malus’s Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Average Intensity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy Transmitted

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Time Period T

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation