76.

(a) We note that d = (76

× 10

6

nm)/40000 = 1900 nm. For the first order maxima λ = d sin θ, which

leads to

θ = sin

−1

λ

d



= sin

−1

589 nm

1900 nm



= 18

◦

.

Now, substituting m = d sin θ/λ into Eq. 37-27 leads to D = tan θ/λ = tan 18

◦

/589 nm = 5.5

×

10

−4

rad/nm = 0.032

◦

/nm. Similarly for m = 2 and m = 3, we have θ = 38

◦

and 68

◦

, and the

corresponding values of dispersion are 0.076

◦

/nm and 0.24

◦

/nm, respectively.

(b) R = N m = 40000 m = 40000 (for m = 1); 80000 (for m = 2); and, 120, 000 (for m = 3).