Question

A diffraction grating contains 5000 lines per nm. If light of wavelength of 5896 Å is incident on it, find the angle of diffraction of the second maxima.

Answer 1

Answer:

$\theta=17.1452^{\circ}$

Explanation:

For transmission grating, $d \sin \theta=n \lambda$

where is angle of diffraction,

is order of diffraction and is wavelength of incident light

where is number of lines per unit length

we have, $d=1 / N=1 / 5000 \mathrm{cm}$ or $\mathrm{d}=(1 /5000) \times 10^{-2} \mathrm{m}$

Hence we get diffraction angle from eqn.(1).

$(1 / 5000) \times 10^{-2} \sin \theta=5896 \times 10^{-10}$

we get angle $\theta=17.1452^{\circ}$