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Compare the amplitude reflection coefficients for an air-water \(\left(n_{\mathrm{w}}=4 / 3\right)\) interface with that of an air-crown glass \(\left(n_{g}=3 / 2\right)\) interface, both at near-normal incidence. What are the corresponding ratios of the reflected to the incident irradiances?

Short Answer

Expert verified
The ratio of reflected to incident irradiance is 1/49 for air-water and 1/25 for air-crown glass interface.

Step by step solution

01

Understand the amplitude reflection coefficient

The amplitude reflection coefficient \( r \) for light going from medium 1 (air, \(n_1 = 1\)) to medium 2 (with refractive index \(n_2\)) is given by the formula: \[ r = \frac{n_1 - n_2}{n_1 + n_2} \]. We'll use this formula to compute the reflection coefficients for both the air-water and air-crown glass interfaces.
02

Calculate the reflection coefficient for air-water interface

Using \( n_1 = 1 \) (air) and \( n_2 = \frac{4}{3} \) (water), the reflection coefficient \( r \) is calculated as: \[ r_{aw} = \frac{1 - \frac{4}{3}}{1 + \frac{4}{3}} = \frac{-\frac{1}{3}}{\frac{7}{3}} = -\frac{1}{7} \].
03

Calculate the reflection coefficient for air-crown glass interface

Using \( n_1 = 1 \) (air) and \( n_2 = \frac{3}{2} \) (crown glass), the reflection coefficient \( r \) is calculated as: \[ r_{ag} = \frac{1 - \frac{3}{2}}{1 + \frac{3}{2}} = \frac{-\frac{1}{2}}{\frac{5}{2}} = -\frac{1}{5} \].
04

Reflectance and corresponding ratios of reflected to incident irradiances

Reflectance \( R \) is the square of the magnitude of the reflection coefficient. For air-water: \( R_{aw} = \left(-\frac{1}{7}\right)^2 = \frac{1}{49} \). For air-crown glass: \( R_{ag} = \left(-\frac{1}{5}\right)^2 = \frac{1}{25} \). Thus, the ratios of reflected to incident irradiances at these interfaces are \( \frac{1}{49} \) and \( \frac{1}{25} \) respectively.

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Key Concepts

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

Air-Water Interface
In optics, an interface refers to the boundary between two different media where the light changes its propagation characteristics. When we talk about the "air-water interface," we're looking at how light behaves as it moves from air, which has a refractive index of approximately 1, into water with a refractive index of 4/3. This change causes some of the light to be reflected back into the air.
The reflection at this interface can be quantified using the amplitude reflection coefficient. At a near-normal angle of incidence, the formula for this coefficient simplifies to:
  • For the air-water interface:
  • The reflection coefficient is calculated by:
\[ r_{aw} = \frac{1 - \frac{4}{3}}{1 + \frac{4}{3}} = -\frac{1}{7} \] This means that a modest fraction of light is reflected, while the majority is transmitted into the water.
Air-Crown Glass Interface
Just like the air-water interface, the "air-crown glass interface" involves light moving from air (n鈧 = 1) to crown glass, which typically has a refractive index of about 3/2. Crown glass is a common optical material used in lenses and various optical components due to its high clarity and quality.
At the interface, light reflects and refracts based on the change in speed as it enters a denser medium, such as crown glass. The amplitude reflection coefficient at normal incidence is again used to determine how much light is reflected:
  • The calculation for the air-crown glass interface is:
\[ r_{ag} = \frac{1 - \frac{3}{2}}{1 + \frac{3}{2}} = -\frac{1}{5} \] This coefficient suggests that a bit more light is reflected at the glass interface compared to water, due to the larger difference in refractive indices.
Refractive Index
The refractive index is a crucial concept in understanding light propagation through different materials. It quantifies how much light slows down in a medium compared to a vacuum (where the speed of light is at its maximum).
The refractive index (n) of air is about 1, indicating minimal slowdown, whereas water has an index of approximately 4/3. Crown glass, on the other hand, has an index of 3/2, indicating a greater reduction in light speed.
  • Higher refractive indices mean that light travels slower in the medium.
  • Different materials bend light to various extents due to differing refractive indices.
Refractive indices help determine the extent of reflection or transmission at an interface. Greater differences between refractive indices at a boundary result in greater reflections, as seen with the air-crown glass interface.
Reflectance
Reflectance is the measure of how much light is reflected by a surface compared to the total incident light. It's represented as a ratio, often expressed as a percentage. Understanding reflectance is important for applications where brightness or minimization of glare is key, such as in solar panels or sunglasses.
The reflectance (R) at an interface is the square of the reflection coefficient's magnitude. Mathematically, this is given as:
  • For air-water:
\[ R_{aw} = \left(-\frac{1}{7}\right)^2 = \frac{1}{49} \]
  • For air-crown glass:
\[ R_{ag} = \left(-\frac{1}{5}\right)^2 = \frac{1}{25} \] This means less light is reflected off the air-water interface compared to the air-crown glass, making water surfaces less reflective than glass.
Irradiance Ratio
The irradiance ratio is a concept closely tied with reflectance. It tells us the proportion of light energy that gets reflected compared to what initially hits a surface. This ratio helps in understanding how surfaces affect light, which is crucial when designing systems like lighting environments or optical devices.
  • The ratio for the air-water interface is:
\[ \text{Irradiance ratio for air-water} = \frac{1}{49} \]
  • For the air-crown glass interface:
\[ \text{Irradiance ratio for air-crown glass} = \frac{1}{25} \] Thus, reflecting more light makes crown glass brighter at a given angle compared to water, which is an important factor when choosing materials for optical clarity and brightness control.

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Most popular questions from this chapter

We know that \(1000 \mathrm{W} / \mathrm{m}^{2}\) of unpolarized light is incident in air on an air-glass interface where \(n_{t i}=3 / 2 .\) If the transmittance for light with its \(E\) -field perpendicular to the plane of incidence is \(0.80,\) how much of that light is reflected?

A beam of light from an argon laser \(\left(\lambda_{0}=500 \mathrm{nm}\right)\) traveling in a glass block \(\left(n_{g}=3 / 2\right)\) is totally internally reflected at the flat airglass interface. If the beam strikes the interface at \(60.0^{\circ}\) to the normal, how deep will the light penetrate into the air before its amplitude drops to about \(36.8 \%\) of its value at the interface?

Imagine that we have a nonabsorbing glass plate of index \(n\) and thickness \(\Delta y,\) which stands between a source \(S\) and an observer \(P\). (a) If the unobstructed wave (without the plate present) is \(E_{u}=\) \(E_{0} \exp i \omega(t-y / c),\) show that with the plate in place the observer sees a wave $$E_{p}=E_{0} \exp i \omega[t-(n-1) \Delta y / c-y / c]$$ (b) Show that if either \(n \approx 1\) or \(\Delta y\) is very small, then $$E_{p}=E_{u}+\frac{\omega(n-1) \Delta y}{c} E_{u} e^{-i \pi / 2}$$ The second term on the right may be envisioned as the field arising from the oscillators in the plate.

Show that \(\tan \theta_{p}=n_{t} / n_{i}\) and calculate the polarization angle for external incidence on a plate of crown glass \(\left(n_{g}=1.52\right)\) in air.

Light of wavelength \(600 \mathrm{nm}\) in vacuum enters a block of glass where \(n_{g}=1.5 .\) Compute its wavelength in the glass. What color would it appear to someone embedded in the glass (see Table 3.4 )?

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