Distinguish between an X-ray image of a body structure and a CT scan.

Question 7

(a) Distinguish between an X-ray image of a body structure and a CT scan.

X-ray image:

CT scan:

[5]

(b) Data for the linear absorption (attenuation) coefficient μ of X-ray radiation of energy 80 keV are given in Fig. 11.1.

metal                           μ / mm^-1

aluminium                   0.46

copper                         0.69

Fig. 11.1

A parallel X-ray beam is incident on a copper filter, as shown in Fig. 11.2.

Fig. 11.2

The intensity of the incident beam is I0.

(i) Calculate the thickness of copper required to reduce the intensity of the emergent beam to 0.25 I0. [2]

(ii) An aluminium filter of thickness 2.4 mm is now placed in the X-ray beam, together with the copper filter in (i).

Calculate the fraction of the incident intensity that emerges after passing through the two

filters. [2]

(iii) Express your answer in (ii) as a gain in decibels (dB). [3]

Reference: Past Exam Paper – June 2014 Paper 42 Q11

Solution:

(a)

An X-ray image is a flat / shadow / 2D image regardless of the depth of the object.

The image from a CT scan is built up from (many) images at different angles. The image is three-dimensional and the image can be rotated (and viewed at different angles).

(b)

(i)

I = I₀ exp(-μx)

0.25I₀ = I₀ exp(-0.69x)

{Since the beam is incident on the copper filter, we need to consider the linear attenuation coefficient of copper.

0.25 = exp (-0.69x)

ln 0.25 = - 0.69 x

Thickness x = ln 0.25 / -0.69}

Thickness, x = 2.0 mm

(ii)

{For a beam incident on a medium of thickness x₁ and linear attenuation coefficient μ₁, the transmitted intensity I is given by

I = I₀ exp(-μ₁x₁)

Fraction of incident intensity that emerges: I / I₀ = exp(-μ₁x₁)

For a beam incident on 2 media, one of thickness x₁ and linear attenuation coefficient μ₁, and the other of thickness x₂ and linear attenuation coefficient μ₂, the transmitted intensity I is given by

I = I₀ exp(-μ₁x₁) × exp(-μ₂x₂)

Fraction of incident intensity that emerges: I / I₀ = exp(-μ₁x₁) × exp(-μ₂x₂)

From the previous part,

For copper, I = 0.25I₀             giving I / I₀ = exp(-μ₁x₁) = 0.25}

For aluminium, I / I₀ = exp (-0.46×2.4) = 0.33

{I / I₀ = exp(-μ₁x₁) × exp(-μ₂x₂)}

Fraction of incident intensity = 0.33 × 0.25 = 0.083

(iii)

{When expressing an answer in dB, we need to take the log of the fraction.}

Gain (in dB) = 10 lg(I/I₀)

Gain (in dB) = 10 lg(0.083)

Gain (in dB) = (-) 10.8 dB

Fig. 4.1 shows the variation with time t of the displacement x of two progressive waves P and Q passing the same point.

Question 19

(a) For a progressive wave, state what is meant by

(i) the period, [1]

(ii) the wavelength. [1]

(b) Fig. 4.1 shows the variation with time t of the displacement x of two progressive waves P and Q passing the same point.

Fig. 4.1

The speed of the waves is 20 cm s^-1.

(i) Calculate the wavelength of the waves. [2]

(ii) Determine the phase difference between the two waves. [1]

(iii) Calculate the ratio

intensity of wave Q

intensity of wave P

[2]

(iv) The two waves superpose as they pass the same point. Use Fig. 4.1 to determine the resultant displacement at time t = 0.45 s. [1]

[Total: 8]

Reference: Past Exam Paper – June 2018 Paper 21 Q4

Solution:

(a)

(i) The period is the time for one oscillation.

(ii) The wavelength is the minimum distance between two points in phase (having the same displacement and moving in the same direction).

(b)

(i)

v = λ / T                       or v = f λ  and f = 1 / T

{λ = v × T

From the graph, the period is 0.60 s.}

λ = 20 × 0.60

λ = 12 cm

(ii)

{Consider the initial position of wave P (at equilibrium position). This is point (0,0).

Wave Q only reaches the equilibrium position at time t = 0.20 s.

The difference between the 2 waves is 0.20 s.

1 period corresponds to a phase difference of 360°. We need to find the phase difference that corresponds to a time of 0.20 s.

0.60 s - - > 360°

0.20 - - > 360° × (0.20 / 0.60)

Alternatively, (since we do not know the direction in which the waves are moving {left or right}) we could say that the ‘trough’ of wave Q occurs before that of wave P. The difference in time is 0.40 s.

Here, we find the phase difference corresponding to 0.40 s.}

phase difference = 360° × (0.20 / 0.60)                     or 360° × (0.40 / 0.60)

phase difference = 120°                                             or 240°

(iii)

{Intensity is proportional to (amplitude)².}

I ∝ A²  

{I_P ∝ 3²

I_Q ∝ 2²}

IQ / IP = AQ² / AP²

IQ / IP = 2.0² / 3.0²

IQ / IP = 0.44

(iv)

{Displacement is a vector quantity, so we need to consider the direction.

From the graph, at t = 0.45 s

Displacement of point on wave P = – 3.00 mm

Displacement of point on wave Q = 1.00 mm}

displacement = 1.00 – 3.00

displacement = –2.00 mm