Question

          A plane diffraction grating is illuminated by a source that emits two spectral lines of wavelengths 420nm and 600nm. Show that the 3rd order of the line of one of these wavelengths is diffracted through a greater a greater angle than the 4th order of the other wavelength.

Answer 1

Answer

$\lambda_1=420nm\\\lambda_2=600nm$

n₁=3

n₂=4

Using bragg equation

$2dsin\theta=n\lambda$

Here interplaner distance and wavelength is different.

So angle relationship

$\frac{sin\theta_1}{sin\theta_2 }=\frac{n_1\lambda_1}{n_2 \lambda_2 }=\frac{3\times420}{4\times600}$

${sin\theta_1}=0.525\times sin\theta_2$

Which means $\theta_1$ Is lesser.