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

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Enhance your skills for the Radiologic Technology Supervisor and Operator Test. Study effectively with multiple choice questions, each supported by explanations and hints to ensure you're fully prepared!

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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