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The critical angle for total internal reflection in a certain media boundary is \(45^{\circ} .\) What is the polarizing (Brewster) angle for light externally incident on the same boundary?

Short Answer

Expert verified
The Brewster angle is approximately \(35.26^{\circ}\).

Step by step solution

01

Understand the Critical Angle

The critical angle is the angle of incidence above which total internal reflection occurs. For this exercise, the critical angle is given as \( 45^{\circ} \). This implies that when light tries to move from the denser to the rarer medium at this angle, it doesn't refract but reflects entirely.
02

Use Snell's Law for the Critical Angle

Snell's Law states that \( n_1 \sin(\theta_c) = n_2 \sin(90^{\circ}) \). Given the critical angle \( \theta_c = 45^{\circ} \), we have \( n_1 \sin(45^{\circ}) = n_2 \cdot 1 \). This simplifies to \( n_1 \sin(45^{\circ}) = n_2 \). Recall \( \sin(45^{\circ}) = \frac{\sqrt{2}}{2} \), hence \( n_1 \frac{\sqrt{2}}{2} = n_2 \).
03

Determine Refractive Indices Relation

From Step 2, \( n_1 \frac{\sqrt{2}}{2} = n_2 \) can be rearranged to \( n_2 = n_1 \frac{\sqrt{2}}{2} \). This relates the two refractive indices across the boundary. Let's denote this as equation (1) for further use.
04

Recall Brewster's Angle Formula

The Brewster angle, \( \theta_B \), is found using the formula \( \tan(\theta_B) = \frac{n_2}{n_1} \). The Brewster angle corresponds to the angle of incidence where reflected light is perfectly polarized.
05

Substitute and Solve for Brewster Angle

Substitute the expression for \( n_2 \) from equation (1) into Brewster's formula: \( \tan(\theta_B) = \frac{n_1 \frac{\sqrt{2}}{2}}{n_1} = \frac{\sqrt{2}}{2} \). Simplifying this, we want \( \theta_B \) such that \( \tan(\theta_B)=\frac{\sqrt{2}}{2} \).
06

Calculate Brewster Angle

The angle \( \theta_B \) where \( \tan(\theta_B) = \frac{\sqrt{2}}{2} \) is \( \theta_B \approx 35.26^{\circ} \). This is derived from the inverse tangent function: \( \theta_B = \tan^{-1}\left(\frac{\sqrt{2}}{2}\right) \approx 35.26^{\circ} \).

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Key Concepts

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

Critical Angle
The critical angle is crucial when discussing light waves and media boundaries. It is the specific angle of incidence above which total internal reflection occurs, meaning light will not refract into the second medium but instead reflect back into the original medium. This phenomenon happens only when light tries to pass from a denser medium (higher refractive index) to a less dense medium (lower refractive index).
The formula to determine the critical angle involves Snell's Law and is given by:
  • \( \sin(\theta_c) = \frac{n_2}{n_1} \)
where \( \theta_c \) is the critical angle, \( n_1 \) and \( n_2 \) are the refractive indices of the denser and rarer medium, respectively. For our problem, the critical angle is stated as \(45^{\circ}\). This means at this angle, light reflects entirely when moving to a less dense medium.
Brewster's Angle
Brewster's Angle is an interesting concept where light undergoes reflection and polarization. It is the angle of incidence at which light reflecting off a surface has maximum polarization, meaning the reflected light is purely polarized parallel to the interface. At this angle, the refracted and reflected light rays are perpendicular to each other.

The formula to calculate Brewster's Angle is:
  • \( \tan(\theta_B) = \frac{n_2}{n_1} \)
where \( \theta_B \) is Brewster's angle, \( n_1 \) is the refractive index of the medium where the incident light is coming from, and \( n_2 \) is the refractive index of the second medium. In the case of the given exercise, we find Brewster's Angle using the relation derived earlier and end up with \( \theta_B \approx 35.26^{\circ} \). This highlights how intertwined reflection, refraction, and polarization are at this specific angle.
Snell's Law
Snell's Law is fundamental in optics for determining how light bends, or refracts, when passing through different media. The law is mathematically expressed as:
  • \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \)
where \( \theta_1 \) and \( \theta_2 \) are the angles of incidence and refraction, respectively, and \( n_1 \) and \( n_2 \) are the refractive indices of the respective media.

In the earlier solution, Snell's Law was utilized to understand the critical angle. For a given critical angle \( \theta_c \), Snell’s Law simplifies to ensure that beyond this angle, total internal reflection occurs. This principle ties together much of what we see in refraction, enabling calculations of angles such as Brewster's Angle through understanding refractive indices.
Total Internal Reflection
Total Internal Reflection describes the complete reflection of a light ray traveling within a medium when it hits a boundary and attempts to move into a less dense medium at an angle greater than the critical angle. Instead of refracting, the light is entirely reflected back into the original medium.

Some conditions need to be met for total internal reflection:
  • The light travels from a denser medium to a rarer medium.
  • The angle of incidence should be greater than the critical angle.
In optical fibers, for innovation in communication technology, this principle is employed to trap light within the core of the fiber. This fundamental concept in wave optics shows how we can harness light effectively in practical applications, from data transmission to creating visually striking optical effects.

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Most popular questions from this chapter

The angle of incidence is adjusted so there is maximum linear polarization for the reflected light from a transparent piece of plastic in air. (a) There is (1) no, (2) maximum, or (3) some light transmitted through the plastic. Explain. (b) If the index of refraction of the plastic is 1.40 , what would be the angle of refraction in the plastic?

When unpolarized light is incident on a polarizer-analyzer pair, \(30 \%\) of the original light intensity passes the analyzer. What is the angle between the transmission axes of the polarizer and analyzer?

Show that when the reflected light is completely polarized, the sum of the angle of incidence and the angle of refraction is equal to \(90^{\circ}\).

(a) Only a limited number of maxima can be observed with a diffraction grating. The factor(s) that limit(s) the number of maxima seen is (are) (a) (1) the wavelength, (2) the grating spacing, (3) both. Explain. (b) How many maxima appear when monochromatic light of wavelength 560 nm illuminates a diffraction grating that has 10000 lines \(/ \mathrm{cm},\) and what are their order numbers?

A film of index of refraction of 1.4 and thickness of \(1.2 \times 10^{-5} \mathrm{~m}\) is on a lens with an index of refraction of 1.6. Light of wavelength \(600 \mathrm{nm}\) is incident normally from air to the film. Consider only reflections from the top and bottom surfaces of the film. (a) How many reflected waves will experience the \(180^{\circ}\) phase shift? (b) What is the path length difference between the two reflected waves? (c) Will the reflected waves interfere constructively or destructively?

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