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Chapter 33: Problem 32

A polarizer and an analyzer are oriented so that the maximum amount of light is transmitted. To what fraction of its maximum value is the intensity of the transmitted light reduced when the analyzer is rotated through (a) \(22.5^{\circ} ;(\mathrm{b}) 45.0^{\circ} ;(\mathrm{c}) 67.5^{\circ} ?\)

Short Answer

Expert verified
(a) 85.36%, (b) 50%, (c) 14.64%

Step by step solution

01

Understand Malus's Law

Malus's Law states that when polarized light passes through an analyzer, the intensity of the transmitted light \(I\) is given by \(I = I_0 \cos^2 \theta\), where \(I_0\) is the initial intensity of the light and \(\theta\) is the angle between the light's initial polarization direction and the axis of the analyzer.
02

Calculate intensity for \(\theta = 22.5^{\circ}\)

Using Malus's Law, substitute \(\theta = 22.5^{\circ}\). The fraction of the transmitted intensity is \( \cos^2(22.5^{\circ}) = \left(\cos 22.5^{\circ}\right)^2 \). Calculate the cosine value and then square it.
03

Simplify and Compute for 22.5掳

\( \cos 22.5^{\circ} \approx 0.9239 \). Thus, \( \cos^2 22.5^{\circ} = 0.9239^2 \approx 0.8536 \). This means for \(22.5^{\circ}\), the transmitted light intensity is reduced to approximately 85.36% of its maximum value.
04

Calculate intensity for \(\theta = 45^{\circ}\)

Now substitute \(\theta = 45^{\circ}\) into Malus's Law. The fraction of the transmitted intensity is \( \cos^2(45^{\circ}) = \left(\cos 45^{\circ}\right)^2 \). Calculate the cosine and square it.
05

Simplify and Compute for 45掳

\( \cos 45^{\circ} \approx 0.7071 \). Thus, \( \cos^2 45^{\circ} = 0.7071^2 \approx 0.5 \). This means for \(45^{\circ}\), the transmitted light intensity is reduced to 50% of its maximum value.
06

Calculate intensity for \(\theta = 67.5^{\circ}\)

Substitute \(\theta = 67.5^{\circ}\) into Malus's Law. The fraction of the transmitted intensity is \( \cos^2(67.5^{\circ}) = \left(\cos 67.5^{\circ}\right)^2 \). Calculate this value.
07

Simplify and Compute for 67.5掳

\( \cos 67.5^{\circ} \approx 0.3827 \). Thus, \( \cos^2 67.5^{\circ} = 0.3827^2 \approx 0.1464 \). This means for \(67.5^{\circ}\), the transmitted light intensity is reduced to approximately 14.64% of its maximum value.

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Key Concepts

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

Polarizer and Analyzer
In the study of light, polarizers and analyzers play a crucial role in controlling the passage of light waves. A polarizer is a device that selects waves of light that oscillate in a particular direction, effectively filtering out other directions.
When unpolarized light passes through a polarizer, it emerges as polarized light, oscillating along a single plane.
An analyzer, on the other hand, is a secondary polarizer that can be adjusted to either allow this polarized light to pass through or block it, depending on its orientation. The orientation of the analyzer relative to the polarizer is key in determining how much light is transmitted. If both are perfectly aligned, the maximum amount of light will pass through. However, rotating the analyzer changes the angle between the light's polarization direction and the analyzer's axis.
Light Intensity
Light intensity is an expression of the amount of light energy present in a given area. Through the process of polarization, the measured intensity can be controlled.
When dealing with a polarizer and analyzer setup, Malus's Law predicts how the intensity of light that passes through an analyzer changes with its rotation.According to Malus's Law, the intensity of light transmitted through the analyzer, given an initial intensity \(I_0\), is \(I = I_0 \cos^2 \theta\).
Here, \( \theta \) is the angle between the light's initial polarization direction and the axis of the analyzer.
This equation demonstrates the dependency of transmitted light intensity on the cosine of the angle between the polarizer and analyzer.
  • When \( \theta = 0^{\circ} \), the intensity remains at its maximum, \( I = I_0 \)
  • As \( \theta \) approaches \( 90^{\circ} \), \( \cos^2 \theta \) approaches zero, reducing \( I \) significantly
Angle of Rotation
The angle of rotation refers to the angle by which the analyzer is rotated from its initial position aligned with the polarizer.
This angle is pivotal in calculating how much light intensity is reduced due to analyzers' orientation.Consider angles in degrees such as \(22.5^{\circ}, 45^{\circ},\) and \(67.5^{\circ}\):
  • At \(22.5^{\circ} \), the intensity fraction is approximately 85.36% of the maximum intensity, as calculated with \( \cos 22.5^{\circ} \approx 0.9239 \)
  • At \(45^{\circ} \), the intensity is reduced to half, (50%), given \( \cos 45^{\circ} \approx 0.7071 \)
  • Finally, at \(67.5^{\circ} \), the intensity is significantly reduced to around 14.64%, because \( \cos 67.5^{\circ} \approx 0.3827 \)
The rotations significantly impact light intensity by changing the angle \( \theta \) and thus apply Malus's Law in practical scenarios.

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Most popular questions from this chapter

(a) Light passes through three parallel slabs of different thicknesses and refractive indexes. The light is incident in the first slab and finally refracts into the third slab. Show that the middle slab has no effect on the final direction of the light. That is, show that the direction of the light in the third slab is the same as if the light had passed directly from the first slab into the third slab. Generalize this result to a stack of \(N\) slabs. What determines the final direction of the light in the last slab?

A glass plate 2.50 \(\mathrm{mm}\) thick, with an index of refraction of \(1.40,\) is placed between a point source of light with wavelength 540 \(\mathrm{nm}\) (in vacuum) and a screen. The distance from source to screen is 1.80 \(\mathrm{cm}\) . How many wavelengths are there between the source and the screen?

Using a fast-pulsed laser and electronic timing circuitry, you find that light travels 2.50 \(\mathrm{m}\) within a plastic rod in 11.5 \(\mathrm{ns}\) . What is the refractive index of the plastic?

A \(45^{\circ}-45^{\circ}-90^{\circ}\) prism is immersed in water. A ray of light is incident normally on one of its shorter faces. What is the minimum index of refraction that the prism must have if this ray is to be totally reflected within the glass at the long face of the prism?

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. 33.53 . 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 8 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.)
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