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.