(a) For a given diffraction grating, does the smallest difference $\mathit{\Delta }\mathit{\lambda }$in two wavelengths that can be resolved increase, decrease, or remain the same as the wavelength increases? (b) For a given wavelength region (say, around 500 nm), is $\mathit{\Delta }\mathit{\lambda }$ greater in the first order or in the third order?

Given a diffraction grating, wavelengths are given.

$\Delta \lambda$ is smallest difference in two wavelengths.

Resolving power R of grating is defined as

$R=\frac{{\lambda }_{avg}}{\Delta \lambda }$

Here, ${\lambda }_{avg}$is the mean wavelength of two emission lines

and $\Delta \lambda$is wavelength difference between them.

Dispersion of a grating at an angle $\theta$is given by

$\frac{\Delta \theta }{\Delta \lambda }=\frac{m}{d\cos \theta }$

Here, m is order,

d is grating space and

$\Delta \lambda$ is wavelength difference.

(a)

Now, Resolving power R of grating is defined as

$R=\frac{{\lambda }_{avg}}{\Delta \lambda }$.

If the resolving power R remains constant, then as the wavelength, $\lambda$, increases, the difference in wavelength, $\Delta \lambda$, must also increase.

So, $\Delta \lambda$ increases.

(b)

Dispersion of a grating at an angle $\theta$is given by

$\frac{\Delta \theta }{\Delta \lambda }=\frac{m}{d\cos \theta }$

It can be seen that the order and the change in wavelength is inversely related to the order.

Hence, the change in wavelength is greater in the first order, than in the third order.