In order to find the difference of optical distances between the courses QAQ′, QPQ′, we have to express QP − QA, PQ′ − AQ′. To find the former, we have, if OAQ = φ, AOP = ω,
QP² = u² + 4a²sin²½ω − 4au sin ½ω sin (½ω − φ)
= (u + a sin φ sin ω)² − a² sin²φ sin²ω + 4a sin² ½ω(a − u cosφ).
Now as far as ω4
4 sin² ½ω = sin²ω + ¼sin4ω,
and thus to the same order
QP² = (u + a sin φ sin ω)²
− a cos φ(u − a cos φ) sin²ω + ¼ a(a − u cos φ) sin4 ω.
pose that Q lies on the circle u = a cos φ, the middle term vanishes, and we get, correct as far as ω4,
| QP = (u + a sin φ sin ω) √ {1 + | a² sin² φ sin4ω | }; |
| 4u |
so that