PHYS 0B - HW #5 Fall 005, Solutions by David Pace Equations referenced equations are from Griffiths Problem statements are paraphrased [.] Imagine a prism made of lucite (n.5) whose cross-section is a quarter circle of radius a. As shown below, one flat side of the prism rests on a table, while light is incident normal to the other flat side. There is a region, between point P and point Q, on the table that is not illuminated by light from the prism. Find the distance betwen O and Q (note: O is the lower left corner of the prism, you are not being asked for the distance P Q). O P Q Solution We are concerned with the light rays that exit the prism. Begin by noting that since the original rays are incident perpendicular to the left surface (i.e. normal incidence where θ I 0), the transmission angle of the light into the prism will also be θ T 0. This means that the light rays inside the prism will be parallel to the tabletop, as shown in the figure to the right. This says nothing about the amplitude of these rays inside the prism; for the moment we have only determined their orientation within it. To solve this problem we will need to know the transmission angle of the light rays as they exit the prism. This angle, θ T, may be determined from Snell s law, sin θ T sin θ I n n Eq. 9.00 () where we are given the indices of refraction (since it is not given explicitly, assume that n because the prism is either surrounded by air or vacuum) and need to solve for θ I before calculating θ T. In order to determine θ I we will need to figure out the geometry of the reflection/transmission phenomena occurring as the rays exit the prism. See figure for this explanation.
Figure : An arbitrary light ray exiting the prism. Since this is a semi-circular prism, the normal to the rounded surface is the radial vector originating at O. The length of this vector is the prism radius, a. The angle of incidence is shown where the incident ray ( E I ) strikes the surface. Other angles also equal to θ I are known from basic geometry. The angle of transmission, θ T, is shown for the transmitted ray E T. Note that Snell s law tells us θ T > θ I, which is represented in this figure. Once we know the transmission angle that produces the shortest OQ length, we can use the triangle to the right to determine that length. This is possible because once we know the transmission angle we can solve for the incident angle (or the other way around, if we can ever find the transmission angle it is probably because we have already determined the incident angle). Once the incident angle is known, we can apply the law of cosines to the triangle at right and determine the length in question. The smallest possible value of θ T is zero. This occurs for the bottom most ray in the prism. The bottom ray strikes the interface at normal incidence, meaning that the transmitted ray emerges at normal incidence (θ T 0) and never touches the tabletop. From the figures and Snell s law we know that the other rays emerge from the prism directed down toward the table. There is a maximum value of θ T, however, as required by Snell s law, [ ] θ T sin n sin(θ I ) () n
but the argument of the sin function must be less than or equal to one, n n sin(θ I ) (3).5 sin(θ I) (4) sin(θ I ) 3 (5) Using sin(θ I ) /3 allows us to solve for the largest possible θ T, sin(θ T,max ) 3 sin(θ I,max) (6) 3 3 (7) θ T,max π 90 o (8) The maximum transmission angle is 90 o. Such a limit makes physical sense because a 90 o transmission angle from a circular surface means the transmitted ray is tangent to the surface. A transmission angle greater than this value would mean that the transmitted ray is still in the prism, which is not possible. We now know that 0 θ T π/. The first ray to strike the table will be the one with θ T θ T,max. One conceptual argument for this fact is: the bottom ray is known never to reach the table. The top most ray (at θ T θ T,max ) definitely reaches the table. All of the rays with a transmission angle between these two extremes must strike the table between them. As such, the top most ray is the one hitting the table at the point Q. The ray featuring a transmission angle of 90 o is tangent to the prism surface. That means we have a right triangle in which the length OQ is the hypotenuse. This is why the problem asks for this length instead of P Q. One side of this triangle has length equal to a, and another side whose length L may be written as (using figure ), L OQ sin(θ I ) (9) OQ (0) 3 The length of the hypotenuse in this right triangle is calculated by, ( ) OQ a + 3 OQ () ( 4 ) OQ a () 9 OQ 9a 5 (3) 3a 5 (4).34a (5) 3
[.] The polarization of an electromagnetic wave is determined according to the direction of its electric field. Show that when a wave strikes an interface between two media, the reflected and transmitted waves have the same polarization. Let the polarization vectors of these waves be ˆn T cos(θ T ) ˆx + sin(θ T ) ŷ and ˆn R cos(θ R ) ˆx + sin(θ R ) ŷ for the transmitted and reflected waves respectively. Consider the boundary conditions in this system to show that θ T θ R. Solution Regardless of the possible changes in polarization, this is still a problem of reflection and transmission. Consider the case of normal incidence because that will provide the easiest solution (if you were not sure which type of incidence to use and asked about this problem, then Professor Carter or myself would have told you to use normal incidence). The components of the electric and magnetic fields are everywhere parallel to the interface and we use the following boundary conditions (the following are (iii) and (iv) from Eq. 9.74), E E (6) µ B µ B (7) As usual, set the direction of the incident wave propagation to be in the +ẑ direction. The electric fields for all of the waves may be written in the form, E (z, t) Eo e i(kz ωt)ˆn (8) where ˆn is the polarization vector. Note that all of our other work in reflection/transmission utilized ˆn ˆx (e.g. Eqs. 9.75-9.77). The same method is employed here, now using ˆx ˆn. The incident wave is, E I (z, t) EoI e i(k z ωt)ˆx (9) B I (z, t) v EoI e i(k z ωt)ŷ (0) where I can set the polarization of this wave as I wish. The point of this exercise is to show that the reflected and transmitted rays have the same polarization, the incident wave s polarization does not matter. The reflected wave is written, E R (z, t) EoR e i( k z ωt)ˆn R () B R (z, t) v EoR e i( k z ωt) (cos θ R ˆx + sin θ R ŷ) () EoR e i( k z ωt) ( ẑ ˆn R ) (3) v EoR e i( k z ωt) (sin θ R ˆx cos θ R ŷ) (4) 4
where the cross product is used because the magnetic field must be perpendicular to both the electric field and the direction of propagation. The transmitted wave is, E T (z, t) EoT e i(k z ωt)ˆn T (5) B T (z, t) v EoT e i(k z ωt) (cos θ T ˆx + sin θ T ŷ) (6) EoT e i(k z ωt) (ẑ ˆn T ) (7) v EoT e i(k z ωt) ( sin θ T ˆx + cos θ T ŷ) (8) Condition (5) provides the following (the interface is placed at z 0 so the exponential dependence factors out of all the terms), EoI ˆx+ EoR (cos θ R ˆx + sin θ R ŷ) EoT (cos θ T ˆx + sin θ T ŷ) (9) EoR sin θ R EoT sin θ T (30) where (9) is the y component of (8), the usefulness of which will be shown soon. Condition (6) gives, ( EoI ŷ + ) EoR (sin θ R ˆx cos θ R ŷ) v µ v This time keep the x component terms, where β is given in (5). µ v EoR sin θ R µ v ( ) EoT ( sin θ T ˆx + cos θ T ŷ) µ v EoT sin θ T (3) EoR sin θ R β EoT sin θ T (3) One way to proceed from this point is to subtract (3) from (9). The result is, 0 ( + β) sin θ T (33) This is satisfied when either β or the sine term is zero. Using the definition of β, where the propagation speeds are positive definite, this condition could be met if one of the permeabilities is negative. Such novel materials are studied and have unique properties, but they are not a part of this course and you are not expected to apply the physics of such materials. We generally assume all permeabilities are close to that of free space and certainly not negative. Our solution is sin θ T 0. For transmission we limit the angle to 0 θ T π/ (if the angle was greater than π/ then the wave would be propagating back through the glass instead of being transmitted, just as in problem [.]). This means that we have shown θ T 0. Now, going back to either (9) or (3) and considering that the amplitude terms and β are non-zero we see that sin θ R sin θ T 0 and therefore θ R θ T 0. Replacing this in the equations for all of the original wave equations shows that they all have the same polarization. 5
[3.] Consider the problem of using a submerged light source to illuminate the surface of a body of water. If the light is placed at a depth d m below the surface, determine the surface area of the light seen on the surface. The relative permittivity of water at optical frequencies is ɛ r.77. Solution Treat the light as a point source. A point source will create a circular pattern on the surface of the water because its light is emitted equally well in all directions. The emitted light is then going from water to air, meaning we have a reflection/transmission problem. Solve for the index of refraction of water, n ɛ r Eq. 9.70 (34).77 (35) Figure sets up the geometry of the problem and determines the method for solving. Figure : For a point light source, all of the emitted light share the same origin. A few select rays are shown here to illustrate the process that will determine the area of light seen on the water s surface. At the surface (solid horizontal line) we have a situation where n > n, meaning that θ T > θ I. As we consider light rays that strike the surface further away from the source, we notice that θ I increases and θ T must increase as well. Once θ T 90 o, there is no transmitted light. The position where this happens will determine the illuminated area on the surface. 6
The radius, r, of the illuminated water surface area is given as part of the triangle shown to the left when θ I θ c is the critical angle (zero transmission). This critical angle is solved for using Snell s law, (), when θ T 90 o. sin θ I n n sin θ T (36).77 (37) where I leave this in terms of sin θ I because, sin θ I r a (38) and the length of r is determined by, [ ] r sin θ I a d + r (39) r r ( r sin θ I ) d (40) d (4) 0.77 (4) (43) employing d. The area of the illuminated surface, A, is, A πr (44) 4.08 m (45) [4.] Solve for the reflection (R) and transmission (T ) coefficients for normal incidence of light waves on the interface between two materials with arbitrary permeabilities µ and µ (repeat what was done in lecture, but don t assume µ µ o ). Also show that R + T. Solution Begin with the definition of the coefficients, R I R I I Eq. 9.86 (46) T I T I I Eq. 9.87 (47) The intensities mentioned above are, I ɛve o (48) 7
Define region to contain the incident and reflected waves (the left side of the interface with the incident wave moving to the right) and region to contain the transmitted wave. The reflection coefficient becomes, R ɛ v E or ɛ v E oi (49) E or E oi (50) Now consider the following general relation between the reflected and incident amplitudes, EoR ( ) β EoI Eq. 9.8 (5) + β β µ v µ v (5) If we divide the incident amplitude from both sides in (50) and then square the result, the complex factors will become unity. Therefore, ( EoR EoI ) E or E oi ( ) β (53) + β ( ) β (54) + β Returning to (49) we have solved for the reflection coefficient, R ( ) β (55) + β Following the same method to solve for T, T ɛ v E ot ɛ v E oi (56) ɛ v ɛ v ( EoT E oi ) (57) The amplitude ratio is found from, EoT ( ) EoI (58) + β ( EoT EoI ( EoT E oi ) ) ( ) (59) + β ( ) (60) + β 8
Insert this result into (56). The velocities are related to ɛ and µ according to, T ɛ ( ) v (6) ɛ v + β v ɛµ Eq. 9.68 (6) Use this to simplify the expression for T. ( ) T β (63) + β Making use of, µ µ µ µ ɛv ɛ v v v µv µ v µ v µ v β (64) Expressing everything in terms of β allows us to easily show R + T. R + T ( ) β ( ) + β (65) + β + β ( β) + 4β ( + β) (66) β + β + 4β + β + β (67) [5.] Problem 9.8 from Griffiths (a) Let there be an amount of free charge in a piece of glass. Roughly how long will it take for this charge to entirely migrate to the surface? (b) How thick should you make the silver coatings to protect yourself from 0 0 Hz microwaves? Bear in mind that silver is expensive so you want to balance the protection it affords with a minimal cost. (c) What are the wavelength and propagation speed of MHz radio waves in copper? Compare this to the wavelength and propagation speed of these waves in air (which you may approximate as vacuum). Solution (a) The characteristic time, τ, for free charge to get to the surface of its conductor is, where τ ɛ/σ. ρ f (t) e (σ/ɛ)t ρ f (0) Eq. 9.0 (68) Table 7. in Griffiths gives the resistivity of glass, and the conductivity σ is the inverse of the resistivity. Therefore, σ 0 Ω/m. 9
The permittivity of glass is ɛ ɛ o ɛ r, where ɛ r is the relative permeability. Glass is non-magnetic so we can write its index of refraction as, n ɛ r Eq. 9.70 (69) The index of refraction of glass is given in Griffiths to be n.5 (p. 39). The permittivity of glass is, ɛ ɛ o (.5) 0 (70) The characteristic time we seek, τ 0 0 0 s (7) (b) Since silver is expensive we want to make these coatings as thin as possible while still blocking the microwaves. The skin depth for microwaves in silver is a measure of how far these waves can penetrate into the metal. If the metal is thinner than the skin depth, then the waves will pass completely through it. Find this distance and then make the coating slightly thicker. Skin depth is given by d /κ where, [ ] ɛµ ( σ ) / κ ω + Eq. 9.6 (7) ɛω ω is the angular frequency of the wave and σ is the conductivity of the material into which the wave is propagating. Take the resistivity of silver from Table 7. in Griffiths to find that σ 6.9 0 7. The angular frequency is ω πf π 0 0. We can use ɛ ɛ o 8.85 0. ɛ ω 0.56 (73) Due to (73) we can simplify κ as, σ ɛω. 08 (74) [ ] ɛµ ( σ ) / κ ω (75) ɛω ɛµ [ σ ] / ω (76) ɛω ωµσ (77) The permeability of conductors is approximately that of vacuum. Check Table 6. in Griffiths to verify this. The skin depth is, d ωµσ so the coatings should be made a little thicker than this. 0 6.4 0 7 m (78)
(c) Wavelength, λ, can be written in terms of wave vector as, λ π k (79) In a conductor the wave vector is given by, [ ] ɛµ ( σ ) / k ω + + Eq. 9.6 (80) ɛω As in part (b) we get the conductivity of copper using Table 7.. Also, the permittivity may be approximated as ɛ o. σ 6 0 7 (8) ɛω ɛ o (π 0 6 ) 5.56 0 5 (8) σ ɛω.08 0 (83) We can skip to the same simplifications made in part (a) due to (8). k ωµσ (84) 5390.60 (85) λ 4.08 0 4 m (86) The propagation speed is, v ω k Eq. 9.9 (87) 408.5 m s (88) In vacuum the speed of an electromagnetic wave is always c 3.0 0 8 m/s. The corresponding wavelength is, λ π k πc ω (89) 300 m (90) [6.] Problem 9. from Griffiths Solve for the reflection coefficient of light striking an air-silver interface. Use µ µ µ o, ɛ ɛ o, σ 6 0 7, and ω 4 0 5. Solution
The reflection coefficient is defined according to (49). A relationship between the complex amplitudes is given in, EoR β + β EoI Eq. 9.47 (9) where β is now a complex quantity. β µ v µ ω k Eq. 9.46 (9) k k + iκ Eq. 9.5 (93) Arrange this into a useful form, ( EoR EoI ( EoR E oi ) ) β + β (94) β β + β + β R (95) where the starred variable means complex conjugate. From previous work in problem 9.8 we know that silver is a good conductor and can show that it has the property k κ. Using (76) and (83) we can write, k κ ωµσ (96) β µ v ωµσ µ ω ( + i) (97) µ v σ ωµ ( + i) (98) Now make the numeric substitutions given in the problem to find the prefactor in (97), noting that v v air c. σ 6 0 µ v µ o c 7 ωµ (4 0 5 (99) )µ o 9 (00) β 9( + i) (0)
The reflection coefficient is, R β β + β + β (0) ( ) ( ) 9( + i) 9( i) + 9( + i) + 9( i) (03) 57 + 9 () 59 + 9 () (04) 0.93 (05) This shows that 93% of the light incident on the air-silver interface is reflected. [7.] Consider light traveling from x (in the ˆx direction), incident normally on a plate of glass with thickness a. The plate if parallel to the yz plane, with one face at x 0 and the other at x a. The glass has index of refraction n, and is surrounded by vacuum (n ). The incident wave has its electric field in the ŷ direction. (a) Write expressions for the electric and magnetic fields in the three regions: x < 0, 0 < x < a, and x > a. Note: in the first region there will be an incident and reflected wave, in the second region there will be a transmitted wave and a reflected wave (reflected off of the back surface of the glass), and in the third region there will only be a transmitted wave. (b) Using the appropriate boundary conditions, solve for the transmission coefficient, T (here you are interested in the energy transmitted into the vacuum region beyond the glass). Also solve for the reflection coefficient R (this will be the energy reflected in the first region). (c) Plot T and R as a function of ka, where k is the incident wave vector. Interpret your result physically. In particular, comment on how R can equal zero in this case - isn t there always a reflected wave from the first interface? Solution (a) Write out all of the waves in a general manner. Beginning with the incident and reflected waves in region, extending from x < 0, E I (x, t) EoI e i(k x ωt) ŷ E R (x, t) EoR e i( k x ωt) ŷ B I (x, t) v B R (x, t) v EoI e i(k x ωt) ẑ (06) EoR e i( k x ωt) ẑ (07) In region, from 0 < x < a, I define a right-going (subscript r) and left-going wave (subscript l), E r (x, t) Eor e i(k x ωt) ŷ E l (x, t) Eol e i( k x ωt) ŷ B r (x, t) v B l (x, t) v Eor e i(k x ωt) ẑ (08) Eol e i( k x ωt) ẑ (09) where the right-going wave is the transmitted wave from the x 0 interface and the left-going wave is the wave reflected from the x a interface. 3
In region 3, extending to x > a, there is only the final transmitted wave. E T (x, t) EoT e i(k x ωt) ŷ B T (x, t) v EoT e i(k x ωt) ẑ (0) where the k and v values in this region are the same as in region because they are both vacuum. This is a normal incidence problem so we apply the boundary conditions of (5) and (6). The permeability of glass is approximately equal to that of vacuum, so (6) becomes, B B at z 0 () B B 3 at z a () Applying the boundary conditions at z 0 provides (the exponential dependence factors out as in other similar problems), EoI + EoR Eor + Eol (3) ( EoI ) EoR ( Eor ) Eol v v EoI EoR v ( Eor ) Eol v (4) (5) where the reason for writing (4) (and (7) below) will become apparent soon. At z a only the angular frequency parts of the exponentials factor out when the boundary conditions are applied, Eor e ik a + Eol e ik a EoT e ik a (6) v ( Eor e ik a Eol e ik a ) v EoT e ik a (7) Eor e ik a Eol e ik a v v EoT e ik a (8) The expressions in (), (4), (5), and (7) relate the five amplitudes in this problem. We treat the incident wave amplitude as a known parameter (assuming that we have purposely shot this incident wave at the glass). This set then consists of four equations with four unknowns. Begin by rewriting (7), v v [ Eor e ik a Eol e ik a ] EoT e ik a (9) Now let β v /v and then divide (8) by (5), ( Eor β e ika ) Eol e ik a Eor e ik a + Eol e ik a (0) which removes the transmitted wave dependence. 4
Performing further algebra allows Eol to be written directly in terms of Eor, and to make the algebra easier to write out I will further define, Eol β β + eik a Eor () γ β β + eik a () Place (0) into () and (4) to obtain, EoI + EoR Eor + γ Eor [ + γ] Eor (3) EoI [ Eor EoR β γ ] Eor [ γ] β Eor (4) Divide () by (3) and then solve for EoR, EoI + EoR EoI EoR EoR + γ β( γ) + γ + β(γ ) + γ β(γ ) (5) EoI (6) Place (5) into (), EoI + + γ + β(γ ) + γ β(γ ) EoI ( + γ) Eor (7) Eor + γ β(γ ) EoI (8) Now put (7) into (0), Eol γ Eor (9) [ γ + γ β(γ ) ] EoI Solve for the remaining amplitude by placing (9) and (7) into (5), EoT Eor + γ β(γ ) eik a + γ [ EoT eika + γe ik a e ik a e ik a [ EoI + γ β(γ ) EoI + γ β(γ ) ] With this amplitude we have solved for the fields in all of the regions. ] e ik a (30) (3) (3) (b) The reflection and transmission coefficients are defined in (45) and (46). Since the reflected and transmitted rays were are considering here exist in vacuum, their intensities share the same velocities and permittivities. The solutions here are then given entirely by the ratios of the square amplitudes. 5
For R, R [ ] + γ + β(γ ) (33) + γ β(γ ) ( ) + ( ) β β+ eika + β β β+ eika ( ) + β β+ e ika + β β β+ e ika ( ) (34) + β β+ eika β β β+ eika + β β+ e ika β β β+ e ika where the complex conjugates have been taken for the squared complex terms. Multiplying all of this out gives, R Rewrite k in terms of k using, ( cos(k a)) ( + β) 4 + ( β) 4 ( + β) ( β) cos(k a) (35) and then R is, R k I n n k T Eq. 9.9 (36) k nk (37) ( cos(ank )) ( + β) 4 + ( β) 4 ( + β) ( β) cos(ank ) The transmission coefficient is found in a similar way, ( e ik a + γe ik [ ]) a T e ik (39) a + γ β(γ ) ( e ik a + β [ ]) β+ eika e ik a e ik a + β β+ eika β( β (40) β+ eika ) ( e ik a + β [ ]) β+ e ika e ik a e ik a + β β+ e ika β( β (4) β+ e ika ) ( ) ( ) + β β+ e ik a + β β+ e ik a e ik a e ik (4) a [ ] [ ] + β β+ eik a β( β β+ eik a ) + β β+ e ik a β( β β+ e ik a ) 6β (β + ) [( + β) 4 + (β ) 4 cos(k a)( + β) (β ) ] (38) (43) (44) 6β (β + ) [( + β) 4 + (β ) 4 cos(ank )( + β) (β ) ] (45) (c) We can rewrite β so that it is in terms of only the index of refraction of the glass, β v v c n n c 6 n n (46)
where I have applied n for the vacuum region and n is the index of refraction of the glass. The cases in which R 0 are found when the numerator in (37) is equal to zero. This leads to, ank mπ (47) ak mπ n The plot below is for a case of n.5, which is a commonly used value for glass. The coefficients may not always add to unity because of rounding errors in my approximation. This indicates that purposely making the glass a certain thickness will result in zero reflection. Furthermore, this thickness, a, is related to the wavelength of the light inside the glass. This relation will be discussed further when we consider dipole radiation. (48) 0. 0.5 T 0. R 0.05 0 0.5π 0.5π 0.75π π.5π.5π 7