If the distance from the X-ray source is halved, the exposure rate increases by what factor?

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Multiple Choice

If the distance from the X-ray source is halved, the exposure rate increases by what factor?

Explanation:
Exposure rate follows the inverse square law: intensity is proportional to 1/d^2, where d is the distance from the X-ray source. If you halve that distance, you’re using d/2, so the intensity becomes 1/(d/2)^2 = 4/d^2. That means the exposure rate is four times greater. In other words, halving distance increases exposure by a factor of four. The other options don’t fit this square-distance relationship: doubling distance reduces exposure to one quarter, while eight times would require a different distance change.

Exposure rate follows the inverse square law: intensity is proportional to 1/d^2, where d is the distance from the X-ray source. If you halve that distance, you’re using d/2, so the intensity becomes 1/(d/2)^2 = 4/d^2. That means the exposure rate is four times greater. In other words, halving distance increases exposure by a factor of four. The other options don’t fit this square-distance relationship: doubling distance reduces exposure to one quarter, while eight times would require a different distance change.

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