Question

436 nm light shines on a diffraction grating with 4908 lines/cm. What is the angular separation between the second order maximum and the third order maximum?

Answer 1

Given that :

wavelength of the light, $\lambda$ = 436 nm = 436 x 10^-9 m

diffraction grating, d = 4908 lines/cm = (1 cm / 4908)

d = (1 m / 4.9 x 10⁵)

For the second order maximum, m = 2

using an equation -

d sin $\theta$ = m $\lambda$                                               

or    sin $\theta$ = m $\lambda$ / d

inserting the values in above eq.

sin $\theta$₂ = (2) (436 x 10^-9 m) (4.9 x 10⁵) / (1 m)

sin $\theta$₂ = 4272.8 x 10^-4

$\theta$₂ = sin^-1 (0.42728)

$\theta$₂ = 25.3 degree

For the third order maximum, m = 3

sin $\theta$₃ = m $\lambda$ / d

inserting the values in above eq.

sin $\theta$₃ = (3) (436 x 10^-9 m) (4.9 x 10⁵) / (1 m)

sin $\theta$₃ = 6409.2 x 10^-4

$\theta$₃ = sin^-1 (0.64092)

$\theta$₃ = 39.8 degree

angular separation between the second order maximum and the third order maximum which will be given as ::

$\Delta$$\theta$ = $\theta$₃ - $\theta$₂

$\Delta$$\theta$ = [(39.8) - (25.3)] degree

$\Delta$$\theta$ = 14.5 degree