If the distance between the source and image receptor is doubled, by what fraction is the intensity reduced?

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Multiple Choice

If the distance between the source and image receptor is doubled, by what fraction is the intensity reduced?

Explanation:
The main concept is the inverse square relationship between distance and intensity. In radiography, intensity varies as 1/d^2, so when distance doubles, the intensity becomes (1/2)^2 = 1/4 of its original value. Therefore, the intensity is reduced to one quarter of what it was, i.e., it’s reduced by a factor of four.

The main concept is the inverse square relationship between distance and intensity. In radiography, intensity varies as 1/d^2, so when distance doubles, the intensity becomes (1/2)^2 = 1/4 of its original value. Therefore, the intensity is reduced to one quarter of what it was, i.e., it’s reduced by a factor of four.

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