Imagine that you are doing Lab 15 (projecting the tail feathers from the light box) you measure the object distance to be 20 cm and the image distance to be 80 cm. You use this data to calculate that the focal length of the lens to be....

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Consider the setup from Lab 15: using the lens formula to find the focal length from object distance do and image distance di.
Option 1: 16 cm. This matches the calculation: 1/f = 1/do + 1/di = 1/20 + 1/80 = 4/80 + 1/80 = 5/80 = 1/16, so f = 16 cm. This is the correct result based on the lens equation.
Option 2: 8 cm. If someone incorrectly computed f by adding reciprocals as 1/f = 1/20 + 1/80 and then misinverting, they might get f = 8 cm, but the correct algebra gives f = 16 cm. This option represents a common arithmetic slip rather than the proper application of the formula.
Option 3: 'the focal length of the lens will vary with different objects and images'. Under the thin lens model, the focal length f is a property of the lens itself and does not depend on the chosen do or di for that calculation. This statement is misleading for ideal lenses and ignores the fixed nature of f in the lens equation context.
Option 4: 15 cm. This value does not satisfy 1/f = 1/20 + 1/80; substituting f = 15 yields 1/f ≈ 0.0667, whereas 1/do + 1/di = 0.05 + 0.0125 = 0.0625. The mismatch shows it is not the correct focal length for these measurements.
In summary, the correct focal length derived from do = 20 cm and di = 80 cm is f = 16 cm, while the other options arise from arithmetic mistakes, misinterpretations of the lens model, or inconsistent calculations.

Physics Honors-Henrikson Unit 10 Test part 1

Imagine that you are doing Lab 15 (projecting the tail feathers from the light box) you measure the object distance to be 20 cm and the image distance to be 80 cm. You use this data to calculate that the focal length of the lens to be....

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