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Introduction to Health Physics: Fourth Edition - Ruang Baca FMIPA UB

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where<br />

A = attenuation, dB,<br />

t = shield thickness, inches,<br />

μ = relative (<strong>to</strong> copper) permeability,<br />

σ = relative (<strong>to</strong> copper) conductivity, and<br />

f = frequency, Hz.<br />

W Example 14.29<br />

NONIONIZING RADIATION SAFETY 793<br />

Calculate the reduction in electric field intensity by a shield of copper foil 1-mil<br />

(0.001-inch) thick at a frequency of 27 MHz.<br />

Solution<br />

Substituting the appropriate values from Table 14-22 in<strong>to</strong> Eq. (14.53), we have<br />

A = 3.34 × 0.001 × 1 × 1 × 27 × 10 6 = 17.4dB.<br />

The attenuation in decibels, if the incident radiation level is I0 and the transmitted<br />

intensity is I, is given by<br />

A, db = 10 log I0<br />

. (14.54)<br />

I<br />

I0<br />

I<br />

A 17.4<br />

= log−1 = log−1 = 55.<br />

10 10<br />

The transmitted electric field intensity is thus seen <strong>to</strong> be only 1/55 of the incident<br />

intensity.<br />

For shielding against microwave radiation, it is convenient <strong>to</strong> use wire mesh. The<br />

reduction fac<strong>to</strong>r by which a wire-mesh shield reduces the electric field intensity is<br />

given by<br />

⎧<br />

⎫<br />

I0<br />

I<br />

where<br />

= 1<br />

4<br />

⎪⎨<br />

λ<br />

a<br />

⎪⎩<br />

× <br />

ln<br />

1<br />

0.83e 2πr<br />

a<br />

<br />

2πr<br />

ea <br />

−1<br />

λ = wavelength,<br />

a = center-<strong>to</strong>-center distance between wires,<br />

r = radius of wire,<br />

I0 = incident radiation intensity, and<br />

I = transmitted radiation intensity.<br />

⎪⎭<br />

2<br />

⎪⎬<br />

, (14.55)

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