What is the magnification factor when a 42-inch source-to-image distance is used and there is an object-to-image distance of 10 inches?

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Multiple Choice

What is the magnification factor when a 42-inch source-to-image distance is used and there is an object-to-image distance of 10 inches?

Explanation:
In radiographic geometry, magnification is determined by how far the object sits from the source relative to the distance to the image receptor. The magnification factor equals SID divided by SOD, where SOD is the source-to-object distance. Since OID is the object-to-image distance, SOD = SID − OID. Here, SID = 42 inches and OID = 10 inches, so SOD = 42 − 10 = 32 inches. The magnification factor = 42 / 32 ≈ 1.3125, which rounds to 1.3. So the magnification factor is about 1.3. The other numbers would require different SID or OID values.

In radiographic geometry, magnification is determined by how far the object sits from the source relative to the distance to the image receptor. The magnification factor equals SID divided by SOD, where SOD is the source-to-object distance. Since OID is the object-to-image distance, SOD = SID − OID.

Here, SID = 42 inches and OID = 10 inches, so SOD = 42 − 10 = 32 inches. The magnification factor = 42 / 32 ≈ 1.3125, which rounds to 1.3.

So the magnification factor is about 1.3. The other numbers would require different SID or OID values.

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