Solutions To Basic Fraunhofer Diffraction Problems (2)

(1) Solution:

 600 lines/ mm  means 6000 (600 x 10)  slits in one cm,  each of length d. Therefore: 

d = 0.001m/ 600 = 1.66 x 10 ^-6 m

For 2^nd order: m = 2: sin q₂ =  2l/d  

But: l =   550 nm  =   5.5 x 10^-7 m,  so:

arcsin (2l/d)  =  arcsin [2(5.5 x 10^-7 m)/ 1.6 x 10 ^-6 m] = 0.663

q₂ =  0.72 rad  

q₂ =   41.2^o

(2) Solution:

l₁ =   656.48 nm,     l₂ =   656.30 nm

Then the wavelength difference:

Dl =   0.18 nm,  

And the 'normalized' average wavelength for the line is:

(l₂ -  l₁ )/2 =  (656.48 nm  -  656.30 nm) / 2

l =   656.39  nm

The minimum no. of lines needed for 1st order resolution must be:

N =    l /  m Dl     =   656.39 nm/  (1) 0.18 nm  =  3647

(3) Solution:

Similar to that for (2) except  wavelength difference is given, e.g.

Dl =   0.652 Å ,  

The normal wavelength for the line is:

l =   6200 Å ,  

And the minimum no. of lines needed for 2nd order resolution must be:

N =    l /  m Dl     =   6200 Å/  (2)  0.652 Å  =  1.7 x 10 ⁴