If the source-to-image distance is doubled while keeping other exposure factors constant, what happens to the dose at the image receptor?

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Multiple Choice

If the source-to-image distance is doubled while keeping other exposure factors constant, what happens to the dose at the image receptor?

Explanation:
Doubling the source-to-image distance changes how the beam spreads out, and the dose at the image receptor follows the inverse square law: intensity falls off with the square of the distance. When the distance is doubled, the receptor receives 1/2^2, or one quarter, of the original dose. Since exposure factors like mAs and kVp are kept constant, no additional photons are produced to compensate, so the receptor dose is four times smaller. If you wanted to keep the same receptor exposure with a doubled distance, you’d need to increase the mAs (or adjust other factors) to compensate.

Doubling the source-to-image distance changes how the beam spreads out, and the dose at the image receptor follows the inverse square law: intensity falls off with the square of the distance. When the distance is doubled, the receptor receives 1/2^2, or one quarter, of the original dose. Since exposure factors like mAs and kVp are kept constant, no additional photons are produced to compensate, so the receptor dose is four times smaller. If you wanted to keep the same receptor exposure with a doubled distance, you’d need to increase the mAs (or adjust other factors) to compensate.

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