🦴 X-Ray Attenuation

Final Intensity (I): 0
Percentage blocked: 0%

🔊 Ultrasound Reflection

Muscle ≈ 1.63e6 kg m⁻² s⁻¹
Bone ≈ 7.0e6 kg m⁻² s⁻¹
Reflection Coefficient (R): 0
Percentage Reflected: 0%

Principles of Medical Imaging

Diagnostic imaging relies on the interaction between radiation (or waves) and human tissue. Different tissues absorb or reflect energy differently, creating contrast in the resulting image.

1. X-Ray Attenuation

As X-rays pass through a medium, their intensity decreases exponentially. This is known as the attenuation law.

I = I₀ e-μx

Where μ is the linear attenuation coefficient and x is the thickness. Tissues like bone have a high μ compared to soft tissue, appearing lighter on a film.

2. Ultrasound Interpretation

Ultrasound uses sound waves (typically 1–20 MHz). When a wave hits a boundary between two tissues of different Acoustic Impedance (Z), some energy is reflected.

Z = ρc

The intensity reflection coefficient (R) is given by:

R = Iᵣ / Iᵢ = (Z₂ - Z₁)² / (Z₂ + Z₁)²
Exam Tip: Coupling Gel is used in ultrasound because the acoustic impedance of air is much lower than skin. Without gel, nearly 100% of the ultrasound would be reflected at the air-skin boundary.

Deep Dive: Worked Examples

✅ Example 1: Basic Ultrasound Reflection

Calculate the reflection coefficient R between muscle (Z₁ = 1.7 × 10⁶) and bone (Z₂ = 6.4 × 10⁶).

Step 1: Calculate (Z₂ - Z₁) and (Z₂ + Z₁) Difference = 4.7 × 10⁶, Sum = 8.1 × 10⁶
Step 2: Plug into R Formula R = (4.7 / 8.1)² ≈ (0.580)²
Step 3: Solve R ≈ 0.337 (or 33.7%)

✅ Example 2: X-Ray Half-Value Thickness (HVT)

If the linear attenuation coefficient of a material is 0.5 cm⁻¹, calculate the thickness required to reduce intensity by half.

Step 1: Relate HVT to μ 0.5 = e^(-μx) → ln(0.5) = -μx → x = ln(2) / μ
Step 2: Plug in μ x = 0.693 / 0.5
Step 3: Solve x = 1.39 cm

✅ Example 3: PET Scanner Annihilation

In PET, a positron annihilates with an electron. Calculate the energy of each of the two resulting photons.

Step 1: Use E = mc² for one particle Rest mass m = 9.11 × 10⁻³¹ kg, c = 3.00 × 10⁸ m/s
Step 2: Calculate Joules E = (9.11 × 10⁻³¹) × (3.00 × 10⁸)² = 8.199 × 10⁻¹⁴ J
Step 3: Convert to MeV E_MeV = (8.199 × 10⁻¹⁴) / (1.60 × 10⁻¹³) ≈ 0.511 MeV

✅ Example 4: Acoustic Impedance of Fat

Fat has a density of 920 kg/m³ and the speed of sound in fat is 1450 m/s. Calculate its acoustic impedance.

Step 1: Formula Z = ρc
Step 2: Plug in values Z = 920 × 1450
Step 3: Solve Z = 1.33 × 10⁶ kg m⁻² s⁻¹

✅ Example 5: X-Ray Contrast Improvement

Contrast agent Iodine (μ = 25 cm⁻¹) is used. If 2mm of Iodine is present, calculate the ratio of I/I₀.

Step 1: Convert thickness to cm x = 2 mm = 0.2 cm
Step 2: Apply formula I/I₀ = e^(-25 × 0.2) = e⁻⁵
Step 3: Solve Ratio ≈ 0.0067 (or 0.67% transmission)