Question

For a monochromator with a 0.540 meter focal length equipped with a 2200 gr/mm grating that measures 100m x 100m

(a) Calculate first order reciprocal linear dispersion

(b) A substance has two emission lines at 2024.41 angstrom and 2024.81; is it theoretically possible to just resolve these spectral features using the grating described?

(c) If a 25 micrometer entrance and exit slits are used in order to have sufficient light throughput for a detectable signal, is it possible to baseline resolve the two spectral lines

Answer 1

Ans a) First-order reciprocal linear dispersion, D^-1

D ^-1 = d/nF

D ^-1

Ans B) Resolution needed

Rneeded = λ/∆λ = {(2024.41+2024.81)/2}/(2024.41-2024.81) = 2024.61/0.4

=5061.52

Maximum resolution (R)

R = nN

=1*100*2200 line/mm

=22*10⁴

Therefore it is possible to resolve these spectral lines using the grating described above. Here Rneeded is less than the grating’s resolving power, Hence the lines can be resolved.

Ans c)

Using this slit width will pass some ∆λeff that can be calculated as follows

∆λeff = W D-1 ∆λeff = 25*10-3 mm * 0.84175nm/mm = 0.02104nm (Theoritical value)

However, the ∆λeff for the two lines to be baseline separated can be calculated to be:

∆λeff = (λ2 – λ1)/2

∆λeff = (2024.81-2024.41)/2 = 0.2

slit widths narrower than this theoretical values is necessary to

achieve a desired resolution.

Hence these lines cannot be resolved.