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Chapter 33: Problem 27

Unpolarized light with intensity \(I_{0}\) is incident on two polarizing filters. The axis of the first filter makes an angle of \(60.0^{\circ}\) with the vertical, and the axis of the second filter is horizontal. What is the intensity of the light after it has passed through the second filter?

Short Answer

Expert verified
The intensity of the light after passing through both filters is \(0.375*I_{0}\).

Step by step solution

01

Calculate the Intensity After the First Polarizer

The intensity after the first filter, \(I_{1}\), will be half of the incoming intensity, \(I_{0}\), because the first polarizer is oriented at an angle of 60 degrees from the vertical direction. In general, unpolarized light passing through a polarizer will result in half the initial intensity. So we have \(I_{1} = 0.5 * I_{0}\).
02

Calculate the Intensity After the Second Polarizer

The light after passing through the first polarizer is polarized. The intensity of this light after it passes through the second filter is determined by the law of Malus, which states that the intensity of polarized light after passing through a polarizer is given by the initial intensity times the square of the cosine of the angle between the light's polarization and the direction of the filter's polarization axis. Since the first filter's axis makes an angle of \(60.0^{\circ}\) with the vertical and the second filter's axis is horizontal, then the difference between the polarization directions of the filters is \(90^{\circ} - 60^{\circ} = 30^{\circ}\). So, the final intensity \(I_{2}\) is given by \(I_{2} = I_{1} *\cos^{2}(30^{\circ})\) .
03

Substitute the Known Values and Solve

Now we can substitute the value of \(I_{1}\) from Step 1 into the equation from Step 2: \(I_{2} = 0.5*I_{0}*\cos^{2}(30^{\circ})\). Calculate the value of \(\cos^{2}(30^{\circ})\), which is \(\frac{3}{4}\), and substitute it back into the equation: \(I_{2} = 0.5*I_{0}*\frac{3}{4} = 0.375*I_{0}\). So, the intensity of the light after passing through both polarizers is \(0.375*I_{0}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Malus's Law
Malus's Law is a fundamental principle used in the study of polarized light. It helps us determine how light intensity changes as it passes through polarizing materials. This law states that the intensity of polarized light emerging from a polarizer is proportional to the square of the cosine of the angle between the light's polarization direction and the axis of the polarizer. Mathematically, it is expressed as:\[ I = I_0 \cdot \cos^2(\theta) \]where:
  • \(I\) is the transmitted light intensity,
  • \(I_0\) is the initial light intensity,
  • \(\theta\) is the angle between the light's initial polarization direction and the axis of the polarizer.
This means that the angle \(\theta\) plays a crucial role in determining how much of the light's intensity is reduced. For an angle of \(0^\circ\), all of the light passes through because the alignment is perfect. As the angle increases towards \(90^\circ\), less light passes through and the intensity diminishes.
Unpolarized Light
Unpolarized light is light that vibrates in multiple directions. Unlike polarized light which has oscillations in a single direction, unpolarized light consists of waves vibrating in many planes perpendicular to the direction of propagation. Common sources of unpolarized light include natural sunlight and many artificial light sources like light bulbs. When unpolarized light encounters a polarizing filter, it becomes partially polarized, as only the light oscillating in the direction matched by the filter's axis is allowed to pass through. As a result, the light's intensity is reduced to half, because only one component of the vibration is transmitted through the filter. This condition is why Malus's Law involves intensity calculations starting with unpolarized light. After the first polarizer in any setup, the light invariably becomes polarized, reducing its intensity by half, which becomes the new initial intensity for any subsequent calculations using Malus's Law.
Polarizing Filters
Polarizing filters are essential tools in controlling light polarization. These filters work by allowing only the light waves aligned with their axis to pass through, effectively suppressing all other directions of oscillations. When unpolarized light first hits a polarizing filter, it becomes polarized light, with its intensity halved.

Function in a Multi-Filter System

In experiments involving multiple filters, the orientation of these filter axes significantly influences the resultant light intensity. After the first filter, the light is polarized and its intensity is determined by the initial unpolarized light. Subsequent filters further adjust this intensity as outlined by Malus's Law, depending on the angles between their axes. These filters are utilized in a variety of applications such as reducing glare in photography, enhancing contrast in liquid-crystal displays, and even in scientific research to study the properties of light and related phenomena. Multiple polarizing filters can be combined in setups to intricately control the properties and intensity of light.

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Most popular questions from this chapter

A horizontal, parallel-sided plate of glass having a refractive index of 1.52 is in contact with the surface of water in a tank. A ray coming from above in air makes an angle of incidence of \(35.0^{\circ}\) with the normal to the top surface of the glass. (a) What angle does the ray refracted into the water make with the normal to the surface? (b) What is the dependence of this angle on the refractive index of the glass?

A ray of light traveling in water is incident on an interface with a flat piece of glass. The wavelength of the light in the water is \(726 \mathrm{nm}\) and its wavelength in the glass is \(544 \mathrm{nm}\). If the ray in water makes an angle of \(56.0^{\circ}\) with respect to the normal to the interface, what angle does the refracted ray in the glass make with respect to the normal?

When the sun is either rising or setting and appears to be just on the horizon, it is in fact below the horizon. The explanation for this seeming paradox is that light from the sun bends slightly when entering the earth's atmosphere, as shown in Fig. \(\mathbf{P 3 3 . 5 1 .}\) since our perception is based on the idea that light travels in straight lines, we perceive the light to be coming from an apparent position that is an angle \(\delta\) above the sun's true position. (a) Make the simplifying assumptions that the atmosphere has uniform density, and hence uniform index of refraction \(n,\) and extends to a height \(h\) above the earth's surface, at which point it abruptly stops. Show that the angle \(\delta\) is given by $$ \delta=\arcsin \left(\frac{n R}{R+h}\right)-\arcsin \left(\frac{R}{R+h}\right) $$ where \(R=6378 \mathrm{~km}\) is the radius of the earth. (b) Calculate \(\delta\) using \(n=1.0003\) and \(h=20 \mathrm{~km} .\) How does this compare to the angular radius of the sun, which is about one quarter of a degree? (In actuality a light ray from the sun bends gradually, not abruptly, since the density and refractive index of the atmosphere change gradually with altitude. \(.\)

A block of a transparent solid sits on top of the horizontal surface of a block of glass. A ray of light traveling in the glass is incident on the top surface of the glass at an angle of \(62.0^{\circ}\) with respect to the normal to the surface. The light has wavelength \(447 \mathrm{nm}\) in the glass and \(315 \mathrm{nm}\) in the transparent solid. What angle does the ray that refracts into the transparent solid make with the normal to the surface?

Light with a frequency of \(5.80 \times 10^{14} \mathrm{~Hz}\) travels in a block of glass that has an index of refraction of \(1.52 .\) What is the wavelength of the light (a) in vacuum and (b) in the glass?
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