The mechanical resolution is determined by the minimum angle ΔN by which the viewing telescope can turn. For PHASMATRON's angle measuring devices, this is 10^-3 degrees (a millionth of a degree). Converting to radians, ΔN=1.74532*10^-5 rad. Next we need dE/dλ so we can approximate Δλ.

We know that dE/dn=2/cos{sin^-1(n_D/2)}. This for n_D=1.72803 gives:
dE/dn=3.9724624 rad. (In the area of sodium D), and
dn/dλ=1.2702*10^-5/A. (again in the area of sodium D)
=> dE/dλ=(dE/dn)(dn/dλ)=5.04582*10^-5 rad/A. Therefore we can approximate and use ΔE/Δλ=5.04582*10^-5 rad/A, and since ΔE almost eqs ΔN, Δλ=ΔN/5.04582*10^-5 rad/A. For ΔN=1.74532*10^-5 rad, this gives:

Δλ_mechanical=0.3458942A. (In the area of the D lines)
Compare this with Δλ_optical=0.3866148A. (In the area of the sodium D lines).
(The TRUE mechanical resolution can be calculated using the program in section Measuring Wavelengths. If you input N=60°, M=59.097° and on another run input N=60° and M=59.098°, you can subtract the two wavelength values found and get thus the Δλ. This way you will get that the true mechanical resolution in the area of sodium lines is 0.3564453A, which is very close to the value of Δλ_mechanical found above.)

Suppose you wanted to calculate instead the mechanical resolution in the area of the blue mercury line. (4358.35A) Then:

dn/dλ=(1.76197-1.74805)/|4358.35-4799.9107|=3.1524545*10^-5/A.
dE/dn=2/cos{sin^-1(n_4358.35/2)}=4.22704 rad.
dE/dλ=4.22704 rad*3.1524545*10^-5/A=1.3325551*10^-4 rad/A.
Δλ_mechanical=1.74532*10^-5 rad/1.3325551*10^-4 rad/A=0.131A.
Compare this with Δλ_optical=0.1152104A from Lord Rayleigh's formula Δλ=λ/{B(dn/dλ)}, for dn/dλ=3.1524545*10^-5/A, B=12*10^-8A, and λ=4358.35A.
Observe that the mechanical resolution varies according to area, roughly in the same way that the optical resolution does.

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