Problem 38 - Master Resource Book in JEE Main Physics

Question

A beam of unpolarized light having flux \(10^{-3} \mathrm{~W}\) falls normally on a polarizer of cross-sectional area \(3 \times 10^{-4} \mathrm{~m}^{2}\). The polarizer rotates with an angular frequency of \(31.4 \mathrm{rads}^{-1}\). The energy of light passing through the polarizer per revolution will be (a) \(10^{-4} \mathrm{~J}\) (b) \(10^{-3} \mathrm{~J}\) (c) \(10^{-2} \mathrm{~J}\) (d) \(10^{-1} \mathrm{~J}\)

Step-by-Step Solution

Answer

The energy of light passing through the polarizer per revolution is (a) \(10^{-4} \mathrm{~J}\).

Step 1: Understand Unpolarized Light Flux

The flux of light is given as \(10^{-3} \mathrm{~W}\), which means that this amount of power is incident on the polarizer. This is for one second of exposure to the light.

Step 2: Calculate Light Transmission Through Polarizer

An unpolarized light passing through a single polarizer lets through half its intensity. Hence, the power of the transmitted light is given by \( \frac{1}{2}\times 10^{-3} \mathrm{~W} = 0.5 \times 10^{-3} \mathrm{~W} \).

Step 3: Determine the Period of Rotation

The angular frequency given is \(31.4 \mathrm{rads}^{-1}\). We can find the period of one complete revolution using \( T = \frac{2\pi}{\omega} \). Here, \( \omega = 31.4 \mathrm{rads}^{-1} \), resulting in \( T = \frac{2\pi}{31.4} \approx \frac{6.28}{31.4} \approx 0.2 \mathrm{s} \).

Step 4: Calculate Energy per Revolution

Energy per revolution is the power times the time for one revolution. Substitute the transmitted power (from Step 2) and the period of rotation (from Step 3) into the energy formula: \( E = P \times T = 0.5 \times 10^{-3} \mathrm{~W} \times 0.2 \mathrm{s} \). This results in \( E = 0.1 \times 10^{-3} \mathrm{~J} = 10^{-4} \mathrm{~J} \).

Key Concepts

Unpolarized Light

Unpolarized light is a type of light where the electric field vectors are oriented randomly in all directions perpendicular to the direction of propagation. This means that there is no specific direction or pattern to the oscillations of the electric fields in the light. When unpolarized light passes through a polarizer, which is a special filter that allows light vibrating in only one direction to pass through, its intensity is reduced.

In this exercise, the original beam of unpolarized light has a flux of \(10^{-3} \; \mathrm{W}\), indicating that is how much power is incident on the polarizer. A polarizer typically transmits about half of the incident light's intensity. Thus, when our beam of unpolarized light encounters the polarizer, only half of its original intensity makes it through. This is a critical point in understanding how polarizers affect light and why we only see half of the intensity: the transmitted power is \(0.5 \times 10^{-3} \; \mathrm{W}\).

To sum up, unpolarized light undergoing polarization results in a consistent reduction in intensity. This is essential when analyzing light interactions and calculations for energy transfer through polarizers.

Angular Frequency

Angular frequency is a concept that quantifies how fast an object, such as a rotating polarizer, is turning. It deals with the rate of rotation and is measured in radians per second (\(\mathrm{rads}^{-1}\)). In the context of this exercise, the polarizer rotates with an angular frequency of \(31.4 \; \mathrm{rads}^{-1}\).

Understanding angular frequency helps in determining the time it takes for one complete revolution, known as the period (\(T\)). The formula for calculating the period is \( T = \frac{2\pi}{\omega} \), where \(\omega\) is the angular frequency. Plugging in our values gives \( T = \frac{2\pi}{31.4} \approx 0.2 \; \mathrm{s}\), meaning it takes 0.2 seconds for the polarizer to make a full rotation.

Angular frequency is a fundamental aspect of rotational dynamics and is critical when analyzing systems involving rotation. It ensures that we can calculate the necessary timing for analyzing energy interactions in rotating objects, such as our polarizer in this scenario.

Energy Calculation

Energy calculation involves understanding how much energy is transmitted through a system or an object over a given period of time. In this exercise, the energy calculation hinges on the interplay between the transmitted power and the time duration of one complete revolution of the polarizer.

The transmitted power through the polarizer is known from our previous discussion about unpolarized light, and is \(0.5 \times 10^{-3} \; \mathrm{W}\). With the period of rotation calculated as \(0.2 \; \mathrm{s}\), we can determine the energy passing through both the polarizer and the light per revolution using the formula:

• Energy \(E = P \times T\)

Substituting the values results in

• \(E = 0.5 \times 10^{-3} \; \mathrm{W} \times 0.2 \; \mathrm{s} = 10^{-4} \; \mathrm{J}\).

This calculation highlights the concept of energy transfer in time-dependent rotational systems. Understanding these interactions provides insight into how energy is manipulated by polarizers in rotating systems.