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Chapter 5: Problem 8

Compare the amplitude and intensity reflection coefficients for light normally incident on a smooth water surface (index \(n=1.33\) ) for the two cases of incidence from air to water and from water to air.

Short Answer

Expert verified
The intensity reflection is 2.0% in both directions due to symmetric Fresnel equations.

Step by step solution

01

Understanding Reflection Coefficients

To compare the reflection coefficients, we need to understand that the amplitude reflection coefficient ( ) relates to the fraction of the wave's amplitude that is reflected. The intensity reflection coefficient ( ^2) is the square of the amplitude reflection coefficient, representing the fraction of incident energy reflected. These coefficients change depending on the direction of incidence due to differences in the refractive indices of the media.
02

Applying Fresnel's Equations (Air to Water)

First, consider light incident from air ( _1 = 1")) to water ( _2 = 1.33"). The amplitude reflection coefficient ( ) for normal incidence is calculated using the formula = rac{n_1 - n_2}{n_1 + n_2}. Substituting the given values, we find = rac{1 - 1.33}{1 + 1.33} = -0.141. Thus, the intensity reflection coefficient is ^2 = (-0.141)^2 = 0.020. This means about 2.0% of the light's energy is reflected.
03

Applying Fresnel's Equations (Water to Air)

Next, consider light incident from water ( _1 = 1.33") to air ( _2 = 1"). The amplitude reflection coefficient in this case is calculated similarly: = rac{n_1 - n_2}{n_1 + n_2} = rac{1.33 - 1}{1.33 + 1} = 0.141. The intensity reflection coefficient is ^2 = (0.141)^2 = 0.020. Hence, the amount of energy reflected is again 2.0%.
04

Comparing the Results

Regardless of whether light is incident from air to water or from water to air, the intensity reflection coefficient is the same, at 2.0%. This is due to the symmetric nature of Fresnel鈥檚 equations at normal incidence, where only the direction of the incident wave changes, not the magnitude of the reflection coefficients.

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Key Concepts

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

Amplitude Reflection Coefficient
The amplitude reflection coefficient is a measure of how much of the wave's amplitude is reflected when a wave strikes an interface between two different media. It gives us an idea of how strong the reflected wave will be compared to the incident wave.

Here's how you can calculate it:
  • Use the formula: \( r = \frac{n_1 - n_2}{n_1 + n_2} \)
  • \( n_1 \) and \( n_2 \) are the refractive indices of the two media involved.
  • A negative value indicates a phase change of 180 degrees in the reflected wave.
For instance, when light moves from air (\( n=1 \)) into water (\( n=1.33 \)), we find \( r = \frac{1 - 1.33}{1 + 1.33} = -0.141 \). Conversely, if moving from water to air, \( r = \frac{1.33 - 1}{1.33 + 1} = 0.141 \). The change in sign highlights the change in phase but not the energy reflected.
Intensity Reflection Coefficient
The intensity reflection coefficient represents the proportion of incident energy that is reflected at the boundary between two different media. It is an essential concept for understanding how much of the wave's power or intensity is reflected back at that boundary.

To calculate this coefficient:
  • Square the amplitude reflection coefficient: \( R = |r|^2 \)
  • It is always a positive number, ranging from 0 to 1.
Using our earlier computed amplitude reflection coefficients, both the air-to-water and water-to-air transitions yield an intensity reflection coefficient of \( R = (-0.141)^2 = (0.141)^2 = 0.020 \). This implies that for either direction, approximately 2.0% of the incident light's energy is reflected, showcasing the consistent energy reflection during normal incidence.
Fresnel's Equations
Fresnel's equations are central to understanding how light behaves at the boundary between two different types of media. These equations describe how light is partially transmitted and partially reflected when it meets an interface.

There are separate equations for normal and oblique incidence, and for calculations involving different polarizations of light. However, for simplicity:
  • At normal incidence, Fresnel's equations simplify to the formulas we used for amplitude and intensity reflection coefficients.
  • The equations take into account both refractive indices of the media, making them crucial for calculating reflection and transmission in optical systems.
In our example, whether light is traveling from air into water or from water into air, Fresnel鈥檚 equations give the same intensity reflection coefficient because of the symmetry present at normal incidence. This is why both transitions reflected the same amount of energy despite switching the direction of incidence.

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Most popular questions from this chapter

Overtones in tuning fork. Does your C523.3 tuning fork emit nothing but a 523 -cps sound? Strike the fork against something hard. You should hear a faint high tone in addition to the strong 523 -cps tone. The high tone dies away in two or three seconds. It is a higher mode of the fork and is strongly damped because it involves greater bending of the prongs. What about the note an octave higher, Cl048? This is difficult to listen for because of the presence of the fundamental, C523. To search for it, use a resonating tube. Tune a tube to \(\mathrm{C} 1046\) by tapping it on your head and listening for the octave above C523. (Or simply cut it by "theory," subtracting \(0.6\) of a radius \(R\) for each end from \(\lambda / 2\) to get the length.) Hold the \(\mathrm{C} 523\) fork at the end of the C1046 tube and listen. (Use a tube tuned at C523 as a control. Move the fork back and forth between the C523 and C1046 tubes.)

Termination of waves on a string. (a) Suppose you have a massless dashpot having two moving parts 1 and 2 that can move relative to one another along the \(x\) direction, which is transverse to the string direction \(z\). Friction is provided by a fluid that retards the relative motion of the two moving parts. The friction is such that the force needed to maintain relative velocity \(\dot{x}_{1}-\dot{x}_{2}\) between the two moving parts is \(Z_{d}\left(\dot{x}_{1}-\dot{x}_{2}\right)\), where \(Z_{i} d\) is the impedance of the dashpot. The input (part 1 ) is connected to the end of a string of impedance \(Z_{1}\) stretching from \(z=-\infty\) to \(z=0 .\) The output (part 2 ) is connected to a string of impedance \(Z_{2}\) that extends to \(z=+\infty\). Show that a wave incident from the left experiences an impedance at \(z=0\) which is the same as that it would experience if connected to a "load" consisting of a string stretching from \(z=0\) to \(+\infty\) and having impedance \(Z_{L}\) given by $$ \mathrm{Z}_{L}=\frac{\mathrm{Z}_{\mathrm{d}} \mathrm{Z}_{2}}{\mathrm{Z}_{d}+\mathrm{Z}_{2}}, \quad \text { that is, } \quad \frac{1}{\mathrm{Z}_{L}}=\frac{1}{\mathrm{Z}_{d}}+\frac{1}{\mathrm{Z}_{2}} \text { . } $$ Thus it is as if the dashpot and string 2 were impedances connected "in parallel" and driven by the incident wave. (b) Show that if string \(Z_{2}\) extends only to \(z=\frac{1}{4} \lambda_{2}\), where \(\lambda_{2}\) is the wavelength in medium 2 (assuming we have a harmonic wave with a single frequency), and there is terminated by a dashpot of zero impedance (frictionless), the wave incident at \(z=0\) is perfectly terminated. Show that the output connection of the dashpot at \(z=0\) cannot tell whether it is connected to a string of infinite impedance or is instead connected to a quarter-wavelength string that is "short-circuited" by a frictionless dashpot at \(z=\frac{1}{4} \lambda_{2 \cdot}\) In either ease the output connection remains at rest.

Reflection from glass. A flat slab of glass reflects about 8\% of incident light intensity for normal incidence, \(4 \%\) in intensity being reflected from each surface. An ordinary silvered mirror reflects more than \(90 \%\) of the visible light. Take a mirror and a single clean piece of glass (a microscope slide, for example). Compare the reflection from the mirror and from the slide with the two held close together so you can see both reflections at once. Look at the reflection of a broad light source like an incandescent bulb, or a piece of white paper, or a patch of sky. Compare the reAlectivity of the slide and that of the mirror at near-normal incidence. Now do the same at near- grazing incidence. At near-grazing incidence, the source, mirror reflection, and slide reflection should be nearly indistinguishable; i.e., you get nearly \(100 \%\) reflection at near-grazing ineidence. At near-normal incidence, the glass should be noticeably dimmer than the mirror. Next take four clean microscope slides. Lay them on top of one another in a series of "steps," with the first slide giving a "floor, " the second slide giving the first "step," and the remaining two slides giving a second "step" of double height. Thus, you can compare at the same time reflection at near- normal incidence from one slide, two slides, and four slides. Look at a broad source (the sky) reflected at near-normal incidence. Neglecting the complications of internal reflections, you should transmit about \(0.92\) through each slide. Thus, four slides should transmit \((0.92)^{4}=0.72\) and reflect \(1-(0.92)^{4} \approx 0.28\) Now make a pile of about a dozen clean slides; they should reflect \(1-(0.92)^{12}=0.64\). Compare with the mirror. Suppose the formula keeps working (and the slides are elean). How many slides will equal one good mirror if the mirror reflects \(93 \%\) of the intensity? Try it-compare the stack of slides with the mirror at near-normal incidence. Also look directly at the source through the stack to see the transmitted light. (It takes about 32 slides, according to the formula. Needless to say, they should be free of fingerprints.) (Microscope slides cost about 35 cents per dozen. Three dozen slides is a good investment for home experiments.)

General sinusoidal wave. Write the traveling wave \(\psi(\pi, t)=A \cos (\omega t-k z)\) as a superposition of two standing waves. Write the standing wave \(\psi(z, t)=A \cos \omega t\) \(\cos k z\) as a superposition of two traveling waves traveling in opposite directions. Consider the following superposition of traveling waves: $$ \psi\langle z, t)=A \cos (\omega t-k z)+R A \cos (\omega t+k z) $$ Show that this sinusoidal wave can be written as a superposition of standing waves given by $$ \psi(z, t)=A(1+R) \cos \omega t \cos k z+\Lambda(1-R) \sin \omega t \sin k z $$ Thus the same wave can be thought of as a superposition either of standing waves or of traveling waves.

Reflections in transmission lines. Suppose a coaxial transmission line having 50 ohms characteristic impedance is joined to one having 100 ohms characteristic impedance. (a) A voltage pulse of \(+10\) volts (maximum value) is incident from the \(50 \Omega\) line to the \(100 \Omega\) line. What is the "height" (in volts, inchding the sign) of the reflected pulse? Of the transmitted pulse? (b) A \(+10\) -volt pulse is incident from the \(100 \Omega\) to the \(50 \Omega\) line. What are the reflected and transmitted pulse heights?
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