

Angular separation, θ, is equal to the separation, s, of two sources?divided by the distance, d, between the sources?and the observer
Worked ExampleA student looks at a helicopter in the night sky with one eye closed and can just resolve two lights as individual sources. The wavelength of both sources is 530 nm. The approximate diameter of the student’s pupil is 6.0 mm. The distance from the student to the helicopter is 6.0 km.
Determine the minimum distance between the lights.

A student views a car in the distance. It has headlights which are 1.5 m apart. The wavelength of light from the car headlights is 500 nm and the pupil diameter of the student is 4.0 mm.
Estimate the maximum distance at which the two headlights could be resolved by the student.
Exam TipYou might be curious where the factor of 1.22 comes from, however, the derivation of this is beyond the scope of the IB DP Physics course so just make sure you know how to use it in your calculations
轉載自savemyexams
以上就是關于【IB DP Physics: HL復習筆記9.4.2 Rayleigh Criterion Calculations】的解答,如需了解學校/賽事/課程動態,可至翰林教育官網獲取更多信息。
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