I.E. Irodov Solutions on Dispersion and Absorption Of Light

1. A free electron is located in the field of a monochromatic light wave. The intensity of light is I = 150 W/m², its frequency is ω = 3.4.10¹⁵ s^-1. Find:

(a) the electron’s oscillation amplitude and its velocity amplitude;

(b) the ratio F_m/F_e, where F_m and F_e are the amplitudes of forces with which the magnetic and electric components of the light wave field act on the electron; demonstrate that that ratio is equal to ½ v/c where v is the electron’s velocity amplitude and c is the velocity of light. Instruction. The action of the magnetic field component can be disregarded in the equation of motion of the electron since the calculations show it to be negligible.

In a travelling plane electromagnetic wave the intensity is simply the time-averaged magnitude of the pointing vector:

I =

= √(ε₀/μ₀)E²

= cε₀E²

On using c = 1/√(ε₀μ₀)

E√ε₀ = H√μ₀

(a) Represent the electric field at any point by E = E₀ sin ωt. Then for the electron we have the equation

mx = eE₀ sin ωt

So x = -(eE₀/mω²) sin ωt

The amplitude of the forced oscillation is

eE₀/mω² = (e/mω²) √(2I/cε₀)

= 5.1 × 10^-16 cm

The velocity amplitude is clearly

eE₀/mω = 5.1 × 10^-16×3.4 ×10¹⁵

= 1.73 cm/sec

(b) For the electric force F_e = amplitude of the electric force

= eE₀

For the magnetic force (which we have neglected above), it is

evB = evμ₀H

= evE√(ε₀μ₀)

= evE/c

writing v = -v₀ cosωt

where v₀ = eE₀/mω

we see that the magnetic force is apart from a sign

(ev₀E₀/2c)sin 2ωt

Hence F_m/F_e = Ratio of amplitudes of the two forces

= v₀/2c

= 2.9× 10^-11

This is negligible and justifies the neglect of the magnetic field of the electromagnetic wave in calculating v₀.

2. Find the free electron concentration in ionosphere if its refractive index is equal to n = 0.90 for radio waves of frequency v = 100 MHz.

n² = 1 – n₀e²/ε₀mω²

= 1 – n₀e²/4π²ε₀mv²

Thus n₀ = (4π²v²mε₀/e²)(1 – n²)

= 2.36×10⁷ cm^-3

3. Assuming electrons of substance to be free when subjected to hard X-rays, determine by what magnitude the refractive index of graphite differs from unity in the case of X-rays whose wavelength in vacuum is equal to λ = 50 pm.

For hard x- rays, the electrons in graphite will behave as if nearly free and the formula of the previous problem can be applied. Thus

n² = 1 – n₀e²/ε₀mω²

and n ≅ 1 – n₀e²/2ε₀mω²

on taking the square root and neglecting higher-order terms.

So (n – 1) = -n₀e²/2ε₀mω²

= -n₀e²λ²/8π²ε₀me²

We calculate n₀ as follows: There are 6×6.023×10²³ electrons in 12 gms of graphite of density 1.6 gm/c.c. Thus n₀ = 6 × 6.023 × 10²³/(12/1.6) per c.c

Using the values of other constants and λ = 50×10^-12 metre we get

n-1 = -5.4×10^-7

4. A sounding of dilute plasma by radio waves of various frequencies reveals that radio waves with wavelengths exceeding λ₀ = 0.75 m experience total internal reflection. Find the free electron concentration in that plasma.

In the plasma radio waves with wavelengths exceeding λ₀ are not propagated. We interpret this to mean that the permittivity becomes negative for such waves. Thus

0 = 1 – n₀e²/ε₀mω² If ω = 2πc/λ₀

Hence n₀e²λ₀²/4π²ε₀mc² = 1

Or n₀ = 4π²ε₀mc²/e²λ₀²

= 1.984×10⁹ per c.c

5. Using the definition of the group velocity u, derive Rayleigh’s formula. Demonstrate that in the vicinity of λ = λ’ the velocity u is equal to the segment v’ cut by the tangent of the curve v (λ) at the point λ’ (Fig. below).

By definition u = dω/dk = (d/dk)(vk)

As ω = vk = v + k dv/dk

Now k = 2π/λ or dk = (-2π/λ²)dλ

Thus u = v – λdv/dλ

Its interpretation is the following

(dv/dλ)_{λ = λ’} is the slope of the v – λ curve at λ = λ’.

Thus as is obvious from the diagram v’ = v(λ’) – λ’(dv/dλ)_{λ = λ’} is the group velocity for λ = λ’.

6. Find the relation between the group velocity u and phase velocity v for the following dispersion laws:

Here λ, k, and ω are the wavelength, wave number, and angular frequency.

(a) v = a/√λ, a = constant

Then u = v – λdv/dλ

= a/√λ – λ(-½ aλ^-3/2)

= (3/2).a/√λ

= (3/2) v

(b) v = bk = ωk, b = constant

So ω = bk² and u = dω/dk = 2bk = 2v

(c) v = c/ω², c = constant = ω/k

So ω² = ck or ω = c^1/3k^/3

Thus u = dω/dk = c^1/3(⅓)k^-2/3

= ⅓ ω/k

= ⅓ v

7. In a certain medium the relationship between the group and phase velocities of an electromagnetic wave has the form uv = c², where c is the velocity of light in vacuum. Find the dependence of permittivity of that medium on wave frequency, ε (ω).

We have uv = (ω/k)dω/dk = c²

Integrating we find ω² = A + c²k², A is a constant.

So k = √(ω² – A)/c

And v = ω/k = c/√(1 – A/ω²)

Writing this as c/√(ε(ω)) we get ε(ω) = 1 – A/ω²

(A can be +ve or negative)

8. A train of plane light waves propagates in the medium where the phase velocity v is a linear function of wavelength: v = a + bλ, where a and b are some positive constants. Demonstrate that in such a medium the shape of an arbitrary train of light waves is restored after the time interval T = 1/b.

We write v = ω/k = a + bλ

So ω = k(a + bλ) = 2πb + ak

Since k = 2π/λ. Suppose a wavetrain at time t = 0 has the form

F(x, 0) = ∫f(k)e^ikx dx

Then at time t it will have the form

F(x, t) = ∫f(k)e^{ikx – iωt} dk

= ∫f(k)e^{ikx – i(2πb + ak)t} dk

= ∫f(k)e^{ik(x – at)} e^-i2πbt dk

At t = 1/b = T

F(x, T) = F(x – aT, 0)

So at time t = T the wave train has regained its shape though it has advanced by aT.

9. A beam of natural light of intensity I₀ falls on a system of two crossed Nicol prisms between which a tube filled with certain solution is placed in a longitudinal magnetic field of strength H. The length of the tube is l, the coefficient of linear absorption of solution is x, and the Verdet constant is V. Find the intensity of light transmitted through that system.

On passing through the first (polarizer) Nicol the intensity of light becomes ½ I₀ because one of the components has been cut off. On passing through the solution the plane of polarization of the light beam will rotate by φ = VlH

And its intensity will also decrease by a factor e^-xl. The plane of vibration of the light wave will then make an angle 90⁰ – φ with the principal direction of the analyzer Nicol. Thus by Malus’ law the intensity of light coming out of the second Nicol will be

½ I₀ e^-xl cos²(90⁰ – φ)

= ½ I₀ e^-xl sin² φ

10. A plane monochromatic light wave of intensity l₀ falls normally on a plane-parallel plate both of whose surfaces have a reflection coefficient ρ. Taking into account multiple reflections, find the intensity of the transmitted light if

(a) the plate is perfectly transparent, i.e. the absorption is absent;

(b) the coefficient of linear absorption is equal to x, and the plate thickness is d.

(a) The multiple reflections are shown below. Transmission gives a factor (1 – ρ) while reflections give factors of ρ. Thus the transmitted intensity assuming incoherent light is

(1 – ρ)²I₀ + (1 – ρ)²ρ² I₀ + (1 – ρ)²ρ⁴I_o + …

= (1 – ρ)²I₀(1 + ρ² + ρ⁴ + ρ⁶ + …)

= (1 – ρ)²I₀×1/(1 – ρ²)

= I₀(1 – ρ)/(1 + ρ)

(b) When there is absorption, we pick up a factor σ = e^-xd in each traversal of the plate. Thus we get

(1 – ρ)²σI₀ + (1 – ρ)²σ³ρ²I₀ + (1 – ρ)²σ⁵ρ⁴I₀ +…

= (1 – ρ)²σI₀(1 + σ²ρ² + σ⁴ρ⁴ + …)

= I₀σ(1 – ρ)²/(1 – σ²ρ²)

11. Two plates, one of thickness d₁ = 3 .8 mm and the other of thickness d₂ = 9.0 mm, are manufactured from a certain substance. When placed alternately in the way of monochromatic light, the first transmits T₁= 0.84 fraction of luminous flux and the second, T₂ = 0.70. Find the coefficient of linear absorption of that substance. Light falls at right angles to the plates. The secondary reflections are to be neglected.

We have

where ρ is the reflectivity; see previous problem, multiple reflection have been ignored.

Thus T₁/T₂ = e^{x(d2 – d1)}

Or x = ln(T₁/T₂)/(d₂ – d₁)

= 0.35 cm^-1

12. A beam of monochromatic light passes through a pile of N = 5 identical plane-parallel glass plates each of thickness 1= 0.50 cm. The coefficient of reflection at each surface of the plates is p = 0.050. The ratio of the intensity of light transmitted through the pile of plates to the intensity of incident light is r = 0.55. Neglecting the secondary reflections of light, find the absorption coefficient of the given glass.

On each surface, we pick up a factor (1 – ρ) from reflection and a factor e^-χl due to absorption in each plate.

Thus T = (1 – ρ)^2N e^-χNl

Thus χ = (1/Nl)ln (1 – ρ)^2N/T

= 0.034 cm^-1

13. A beam of monochromatic light falls normally on the surface of a plane-parallel plate of thickness l. The absorption coefficient of the substance the plate is made of varies linearly along the normal to its surface from x₁ to x₂. The coefficient of reflection at each surface of the plate is equal to ρ. Neglecting the secondary reflections, find the transmission coefficient of such a plate.

Apart from the factor (1 – ρ) on each end face of the plate, we shall get a factor due to absorptions. This factor can be calculated by assuming the plate to consist of a large number of very thin slab within each of which the absorption coefficient can be assumed to be constant Thus we shall get a product like

…..e^-χ(x)dx e^{-χ(x + dx)dx} e^{-χ(x + 2dx)dx} …..

This product is nothing but

Now χ(0) = χ₁, χ(l) = χ₂ and variation with x is linear so χ(x) = χ₁ + (x/l)(χ₂– χ₁)

Thus the factor becomes

=

14. A beam of light of intensity I₀ falls normally on a transparent plane-parallel plate of thickness l. The beam contains all the wavelengths in the interval from λ₁ to λ₂ of equal spectral intensity. Find the intensity of the transmitted beam if in this wavelength interval the absorption coefficient is a linear function of λ, with extreme values x₁ and x₂. The coefficient of reflection at each surface is equal to ρ. The secondary reflections are to be neglected.

The spectral density of the incident beam (i.e. intensity of the components whose wavelength lies in the interval λ and λ + dλ) is I₀/(λ₂ – λ₁)dλ, λ₁≤ λ≤ λ₂

The absorption factor for this component is

And the transmission factor due to reflection at the surface is ( 1 – ρ)². Thus the intensity of the transmitted beam is

15. A light filter is a plate of thickness d whose absorption coefficient depends on wavelength λ as x(λ) = α(1 – λ/λ₀)² cm^-1, where α and λ₀ are constants. Find the passband ∆λ of this light filter, that is the band at whose edges the attenuation of light is η times that at the wavelength λ₀. The coefficient of reflection from the surfaces of the light filter is assumed to be the same at all wavelengths.

At the wavelength λ₀ the absorption coefficient vanishes and loss in transmission is entirely due to reflection. This factor is the same at all wavelengths and therefore cancels out in calculating the pass band and we need not worry about it now

T₀ = (transmissivity at λ = λ₀) = (1 – ρ)²

T = (transmissivity at λ ) = (1 – ρ)²e^-χ(λ)d

The edges of the pass band are λ₀ ± ∆λ/2 and at the edge

T/T₀ =

= η

Thus ∆λ/2λ₀ = √((ln 1/η)/αd)

Or ∆λ = 2λ₀ √(1/αd(ln 1/η))

16. A point source of monochromatic light emitting a luminous flux Φ is positioned at the centre of a spherical layer of substance. The inside radius of the layer is a, the outside one is b. The coefficient of linear absorption of the substance is equal to x, the reflection coefficient of the surfaces is equal to ρ. Neglecting the secondary reflections, find the intensity of light that passes through that layer.

We have to derive the law of decrease of intensity in ah absorbing medium taking in to account the natural geometrical fall-off (inverse square law) as well as absorption. Consider a thin spherical shell of thickness dx and internal radius x. Let I(x) and I(x + dx) be the intensities at the inner and outer surfaces of this shell.

Then 4π²I(x) e^{-x dx} = 4π(x + dx)²I(x + dx)²I(x + dx)

Except for the factor e^{-x dx} this is the usual equation. We rewrite this as

x²I(x) = I(x + dx)(x + dx)²(1 + χ dx)

= (I + (dI/dx)dx(x² + 2x dx) (1 + χ dx)

Or x²dI/dx + χ x² I + 2xI = 0

Hence d/dx(x²I) + χ(x²I) = 0

So x²I = Ce^-χx

Where C is the constant of integration.

In our case we apply this equation for a≤ x ≤ b

For x ≤ a the usual inverse square law gives

I(a) = Φ/4πa²

Hence C = (Φ/4π)e^χa

And I(b) = (Φ/4πb²)e^{-χ(b – a)}

This does not take into account reflections. When we do that we get

I(b) = (Φ/4πb²)(1 – ρ)²e^{-χ(b – a)}

17. How many times will the intensity of a narrow X-ray beam of wavelength 20 pm decrease after passing through a lead plate of thickness d = 1.0 mm if the mass absorption coefficient for the given radiation wavelength is equal to μ/ρ = 3.6 cm²/g?

The transmission factor is e^-μd and so the intensity will decrease e^μd

= e^{3.6×11.3×0.1}

= 58. 4 times

We have used μ = (μ/ρ)×ρ and used the known value of density of lead.

18. A narrow beam of X-ray radiation of wavelength 62 pm penetrates an aluminium screen 2.6 cm thick. How thick must a lead screen be to attenuate the beam just as much? The mass absorption coefficients of aluminium and lead for this radiation are equal to 3.48 and 72.0 cm²/g respectively.

We require μ_pbd_pb = μ_Ald_Al

Or (μ_pb/ρ_pb) ρ_pbd_pb = (μ_Al/ρ_Al)ρ_Ald_Al

72.0×11.3×d_pb = 3.48×2.7×2.6

d_pb = 0.3 mm

19. Find the thickness of aluminium layer which reduces by half the intensity of a narrow monochromatic X-ray beam if the corresponding mass absorption coefficient is μ/ρ = 0.32 cm²/g.

1/2 = e^-μd

Or d = ln 2/μ

= ln 2/(μ/ρ)ρ

= 0.80 cm

20. How many 50%-absorption layers are there in the plate reducing the intensity of a narrow X-ray beam η = 50 times?

We require N plates

(½)^N = 1/50

So N = ln 50/ln 2

= 5.6