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Chapter 15: Problem 1

Initially unpolarized light passes in turn through three linear polarizers with transmission axes at \(0^{\circ}, 30^{\circ},\) and \(60^{\circ}\) respectively, relative to the horizontal. What is the irradiance of the product light, expressed as a percentage of the unpolarized light irradiance?

Short Answer

Expert verified
28.125% of the unpolarized light irradiance.

Step by step solution

01

Determine Initial Intensity After First Polarizer

Unpolarized light passing through a polarizer results in half of its initial intensity being transmitted. Thus, if the initial intensity of unpolarized light is \(I_0\), after passing through the first polarizer (with axis at \(0^\circ\)), the intensity becomes \(I_1 = \frac{1}{2}I_0\).
02

Apply Malus's Law for Second Polarizer

The light intensity after the second polarizer can be found using Malus's Law, which states \(I_2 = I_1 \cos^2{\theta}\), where \(\theta = 30^\circ\) is the angle between the axes of the first and second polarizer. Plugging in the values, we get \(I_2 = \frac{1}{2}I_0 \cos^2{30^\circ}\). Calculating, \(\cos{30^\circ} = \frac{\sqrt{3}}{2}\) so \(\cos^2{30^\circ} = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}\). Therefore, \(I_2 = \frac{1}{2}I_0 \times \frac{3}{4} = \frac{3}{8}I_0\).
03

Apply Malus's Law for Third Polarizer

Now apply Malus鈥檚 Law for the third polarizer. The irradiance after the third polarizer is given by \(I_3 = I_2 \cos^2{\theta}\), where \(\theta = 30^\circ\) is the angle between the transmission axes of the second and third polarizers. Using the result from the previous step \(\cos^2{30^\circ} = \frac{3}{4}\), we find \(I_3 = \frac{3}{8}I_0 \times \frac{3}{4} = \frac{9}{32}I_0\).
04

Calculate Final Irradiance as Percentage of Initial

The final step is to express \(I_3\) as a percentage of the original intensity \(I_0\). \(I_3 = \frac{9}{32}I_0\), so the percentage is \(\left(\frac{9}{32}\right) \times 100\% = 28.125\%\).

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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 in optics that describes the intensity of polarized light passing through a polarizer. It states that the irradiance of light after a polarizer is proportional to the square of the cosine of the angle between the light's polarization direction and the polarizer's transmission axis. Here's how it works:
  • The initial light intensity, say, \(I\), reduces to \(I' = I \cos^2{\theta}\) after passing through a polarizer.
  • The angle \(\theta\) is crucial because it represents the orientation difference between the light's polarization and the polarizer's axis.
Malus's Law is indispensable in calculating changes in light intensity through multiple polarizers with different angular orientations. This law helps us predict how strongly light will be attenuated and is essential in practical applications like controlling glare and managing light in scientific instruments.
Linear Polarizers
A linear polarizer is an optical filter that allows light waves of a specific polarization to pass through while blocking waves of other polarizations. When unpolarized light, which contains multiple polarization directions, passes through a linear polarizer, the resulting light is linearly polarized. Here's a closer look:
  • The transmitted light intensity is reduced by at least half鈥攕ince only one polarization direction continues.
  • The polarizer has a carefully oriented filter, blocking all but one direction of vibration.
In multiple polarizer systems, the transmission angle of each polarizer concerning its predecessor can significantly alter the light's behavior. This interplay prepares the foundation for calculations using Malus鈥檚 Law while observing reduced intensities progressively from one polarizer to the next.
Irradiance Calculation
Calculating irradiance involves evaluating the light intensity as it interacts with polarizing elements. In the case of several polarizers aligned at different angles, irradiance decreases stepwise through successive applications of Malus's Law.
  • Initially, for unpolarized light passing through a linear polarizer, the intensity drops to half, as one polarization orientation is selected.
  • For subsequent polarizers, the angle between the transmission axes dictates further intensity reduction.
  • This reduction follows Malus's equation individually for each change in angle.
Using these steps, the combined effect of multiple polarizers can be quantified. This involves calculating with precise angular relationships, ensuring accurate representation of irradiance in terms of an initial light source's percentage. Here, step-by-step application of \(I = I_0 \cos^2{\theta_1} \cos^2{\theta_2} \ldots\) systematically finds the output light's percentage of the starting level.

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Most popular questions from this chapter

In each of the following cases, deduce the nature of the light that is consistent with the analysis performed. Assume a \(100 \%\) efficient polarizer. a. When a polarizer is rotated in the path of the light, there is no intensity variation. With a QWP in front of the rotating polarizer (coming first), one finds a variation in intensity but no angular position of the polarizer that gives zero intensity. b. When a polarizer is rotated in the path of the light, there is some intensity variation but no position of the polarizer giving zero intensity. The polarizer is set to give maximum intensity. A QWP is allowed to intercept the beam first with its OA parallel to the TA of the polarizer. Rotation of the polarizer now can produce zero intensity.

Light from a source immersed in oil of refractive index 1.62 is incident on the plane face of a diamond \((n=2.42),\) also immersed in the oil. Determine (a) the angle of incidence at which maximum polarization occurs and (b) the angle of refraction into the diamond.

The Fresnel equations show that the fraction \(r\) of the incident field that is reflected from a dielectric plane surface for the TE polarization mode has the form, $$r=\frac{\cos \theta-\sqrt{n^{2}-\sin ^{2} \theta}}{\cos \theta+\sqrt{n^{2}-\sin ^{2} \theta}}$$ Thus, the reflectance \(R=r^{2}\) has the form, R=\left(\frac{\cos \theta-\sqrt{n^{2}-\sin ^{2} \theta}}{\cos \theta+\sqrt{n^{2}-\sin ^{2} \theta}}\right)^{2} where \(\theta\) is the angle of incidence and \(n\) is the ratio \(n_{2} / n_{1}\) a. Calculate the reflectance \(R\) for the TE mode when the light is incident from air onto glass of \(n=1.50\) at the polarizing angle. b. The reflectance calculated in part (a) is also valid for an internal reflection as light leaves the glass going into air. This being the case, calculate the net fraction of the TE mode transmitted through a stack of 10 such plates relative to the incident irradiance \(I_{0}\). Assume that the plates do not absorb light and that there are no multiple reflections within the plates. c. Calculate the degree of polarization \(P\) of the transmitted beam, given by $$P=\frac{I_{T M}-I_{T E}}{I_{T M}+I_{T E}}$$ where \(I\) stands for the irradiance of either polarization mode.

Light is incident on a water surface at such an angle that the reflected light is completely linearly polarized. a. What is the angle of incidence? b. The light refracted into the water is intercepted by the top flat surface of a block of glass with index of 1.50 . The light reflected from the glass is completely linearly polarized. What is the angle between the glass and water surfaces?

since a sheet of Polaroid is not an ideal polarizer, not all the energy of the \(\overrightarrow{\mathbf{E}}\) -vibrations parallel to the TA are transmitted, nor are all \(\overrightarrow{\mathbf{E}}\) -vibrations perpendicular to the TA absorbed. Suppose an energy fraction \(\alpha\) is transmitted in the first case and a fraction \(\beta\) is transmitted in the second. a. Extend Malus' law by calculating the irradiance transmitted by a pair of such polarizers with angle \(\theta\) between their TAs. Assume initially unpolarized light of irradiance \(I_{0} .\) Show that Malus' law follows in the ideal case. b. Let \(\alpha=0.95\) and \(\beta=0.05\) for a given sheet of Polaroid. Compare the irradiance with that of an ideal polarizer when unpolarized light is passed through two such sheets having a relative angle between TAs of \(0^{\circ}\) \(30^{\circ}, 45^{\circ},\) and \(90^{\circ}\)
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