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Chapter 38: Problem 45

The critical angle for total internal reflection for sapphire surrounded by air is \(34.4^{\circ} .\) Calculate the polarizing angle for sapphire.

Short Answer

Expert verified
First calculate the refractive index of sapphire. Then, using Brewster's law, obtain the polarizing angle using the calculated refractive index. Finally, convert this angle from radians to degrees to get the final answer.

Step by step solution

01

Calculate the refractive index of Sapphire

We calculate the refractive index of sapphire using the formula: \( n2 = n1 / \sin(\theta_{critical}) \). In this exercise, the n1 is the refractive index of air which is 1 and \(\theta_{critical}\) is the critical angle of total internal reflection for sapphire which is given as \(34.4^{\circ}\). Therefore, \( n2 = 1 / \sin(34.4^{\circ}) \). Let's solve this equation to get the value of n2.
02

Calculate the polarizing angle

After obtaining the refractive index, n2, the polarizing angle can be calculated by the formula: \( \tan(\theta_{polarizing}) = n2 / n1 \). Substitute n2 from Step 1 and n1 = 1 into the formula to calculate the polarizing angle.
03

Convert to degrees

The value of the polarizing angle will be in radian form. Convert it into degrees by using the conversion formula: 1 radian = \(180^{\circ} / \pi\).

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Key Concepts

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

Refractive Index
The refractive index is a measure of how much light slows down as it enters a medium from a vacuum or air. It is denoted by the symbol "n" and is calculated as the ratio of the speed of light in a vacuum to the speed of light in the medium. The formula for refractive index is given by:
  • \( n = \frac{c}{v} \)
where "c" is the speed of light in a vacuum and "v" is the speed of light in the medium.
In this context, Sapphire has a higher refractive index than air, meaning light travels slower in Sapphire than in air. Given the critical angle of sapphire with air is \(34.4^{\circ}\), we can calculate the refractive index of sapphire using the critical angle formula.
By substituting the values, the equation becomes:
  • \( n = \frac{1}{\sin(34.4^{\circ})} \)
Using this formula, we first find the refractive index needed to determine subsequent angles.
Critical Angle
The critical angle is the angle of incidence above which total internal reflection occurs. It is specific to the media鈥檚 refractive indices involved and occurs when light travels from a medium with a higher refractive index to a medium with a lower refractive index.
The formula for finding the critical angle \( \theta_{critical} \) between two mediums is:
  • \( \sin(\theta_{critical}) = \frac{n_2}{n_1} \)
where \( n_1\) is the refractive index of the first medium (here, sapphire) and \( n_2\) is the refractive index of the second medium (air, in this example, with \( n=1\)).
By rearranging, we can find the critical angle when only the refractive indices are known. However, since it is given in the problem as \(34.4^{\circ}\), we used it to find the refractive index of sapphire.
Total internal reflection is a unique phenomenon that enables applications like fiber optics, allowing light to be "trapped" inside cables by repeated reflections.
Polarizing Angle
Also known as Brewster's angle, the polarizing angle is the angle of incidence at which reflected light is completely polarized. This occurs when light hits the boundary of two media, such as sapphire and air, causing intense reflection at a 90-degree angle to the refracted light.
The formula to find this angle is:
  • \( \tan(\theta_{polarizing}) = \frac{n_2}{n_1} \)
Here, \( \theta_{polarizing} \) is Brewster's angle, \( n_2 \) is the refractive index of sapphire, and \( n_1 \) is the refractive index of air (1).
By solving this, we get an angle in radians, which can be converted to degrees for practical use using:
  • \(1 \ \text{radian} = \frac{180^{\circ}}{\pi}\)
Understanding how light polarization works can help in various fields, such as photography and optics, where glare reduction is critical.

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Most popular questions from this chapter

Two wavelengths \(\lambda\) and \(\lambda+\Delta \lambda\) (with \(\Delta \lambda<<\lambda\) ) are incident on a diffraction grating. Show that the angular separation between the spectral lines in the \(m\) th-order spectrum is $$ \Delta \theta=\frac{\Delta \lambda}{\sqrt{(d / m)^{2}-\lambda^{2}}} $$ where \(d\) is the slit spacing and \(m\) is the order number.

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When Mars is nearest the Earth, the distance separating the two planets is \(88.6 \times 10^{6} \mathrm{km} .\) Mars is viewed through a telescope whose mirror has a diameter of \(30.0 \mathrm{cm} .\) (a) If the wavelength of the light is \(590 \mathrm{nm},\) what is the angular resolution of the telescope? (b) What is the smallest distance that can be resolved between two points on Mars?

The scale of a map is a number of kilometers per centimeter, specifying the distance on the ground that any distance on the map represents. The scale of a spectrum is its dispersion, a number of nanometers per centimeter, which specifies the change in wavelength that a distance across the spectrum represents. One must know the dispersion in order to compare one spectrum with another and to make a measurement of (for example) a Doppler shift. Let \(y\) represent the position relative to the center of a diffraction pattern projected onto a flat screen at distance \(L\) by a diffraction grating with slit spacing \(d\). The dispersion is \(d \lambda / d y\). (a) Prove that the dispersion is given by 7 $$ \frac{d \lambda}{d y}=\frac{L^{2} d}{m\left(L^{2}+y^{2}\right)^{3 / 2}} $$ \(x\) (b) Calculate the dispersion in first order for light with a mean wavelength of 550 nm, analyzed with a grating having 8000 rulings \(/ \mathrm{cm},\) and projected onto a screen \(2.40 \mathrm{m}\) away.
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