Question

Please show work for the answer! Thanks in advance
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1) V and Al have emission lines at 308.211nm and 308.215nm respectively. What resolving power would a grating require in order to resolve these emission lines?

2) 25.7mm of an echellette grating is illuminated in order to achieve the resolving power you calculated above. How many lines/mm does this echellette grating have (if n=1st order)?

3) A monochromator has a focal length of 1.6m and a collimating mirror with a diameter of 2.5cm. The dispersing device was a grating with 1000lines/mm. What are the first and second-order reciprocal linear dispersion of the monochromator?

Answer 1

1.

Given : = 308.211 nm ; = 308.215 nm

Solution :

The average wavelength is given by :

= 308.213 nm

The difference in wavelength is given by :

= 308.215 nm - 308.211 nm

= 0.004 nm

The resolvance is given by :

  

= 77053.25

Answer : R = 77053.25