69. In this case the path traveled by ray no. 2 is longer than that of ray no. 1 by 2L/ cos θ
r
, instead of 2L.
Here sin θ
i
/ sin θ
r
= n
2
, or θ
r
= sin
−1
(sin θ
i
/n
2
). So if we replace 2L by 2L/ cos θ
r
in Eqs. 36-34 and
36-35, we obtain
2n
2
L
cos θ
r
=
m +
1
2
λ
m = 0, 1, 2,
· · ·
for the maxima, and
2n
2
L
cosθ
r
= mλ
m = 0, 1, 2,
· · ·
for the minima.