Question

Consider the first order diffraction of 550 nm light from a 600 line/mm grating in a 0.35 meter focal length monochromator.

a) If the incident light is normal to the grating surface, what is the angle of the diffracted light?

b) What is the dispersion at the exit slit? If the grating has a width of 15 cm, what is the diffraction limited resolution (in nm) of the monochromator at 550 nm?

c) What slit width does this correspond to?

d) How do these results change if the monochromator is operated in second order?

Answer 1

Q1

a)

The wavelength of the diffraction light(lamda) =550 nm =450*10^-9 m

Then, calculate lines per meter:

The number of lines on the grating element is (N) =600 line/mm =6000 line/cm =12000*10^2 lines/m

For

The focallength of the lens is (f) =0.25 m

Finally:

The angle of diffraction is given by

sin(ϴ)=nNlamdda =(1)(6000 lines/10^-2m)(550 *10^-9 m)=0.33

Solve for ϴ

Then ϴ=sin-1(0.33) =19.268 ° degrees (angle of diffraction)

b)

The dispersion at the exit slit is given by

Dispersion is given by D =nN/costheta =(1)(600000)/cos(19.268) =635606 rad/m

c)

The diffracted resolution is given by

lamda/dlamda =nN =(1)(6000000*15*10^-2) =90000