Question

The spectrum of sodium has two closely spaced lines, known as the sodium doublet, with wavelengths 589.0 nm and 589.6 nm. When sodium light is incident on a diffraction grating with 4,300 rulings/cm, the maxima corresponding to this doublet are separated by Δy = 4.80 mm when the screen is L = 1.70 m from the grating. What is the value of m in this situation?

Answer 1

We know that

y = m lambda D/d

d = 1/4300 = 2.33 x 10^-4 /cm = 2.33 x 10^-6 per m

m = y d/lambda D

since y2 - y1 is given ; we will take lambda2 - lambda1

m = 4.88 x 10^-3 x 2.33 x 10^-6/(1.7 x (589.6 - 589) x 10^-9) = 11

Hence, m = 11