Question

I JUST NEED #4. I’m supplying my answer for #3 as it is needed.

3. You have access to two diffraction gratins, on with 600 groves/mm and a length of 50 mm and one with 1,200 grooves/mm and a length of 25 mm. Calculate the linear dispersion and resolving power for both. You also plan to use the first-order diffracted beam and use a configuration where B is around 10 degrees and the focal length of the system is 0.3 m 4. Calculate the slit width necessary for a 0.1-nm bandpass for the two grating cases in question 3 and comment on their relative utility in a spectrometer

Answer 1

For this part, we need the reciprocal linear dispersion, as the band pass is roughly given as: (slit width) times Reciprocal linear dispersion almostequalto band pass Therefore, the slit width necessary for a 0.1 nm band pass is slit width = 0.1 nm/Reciprocal linear dispersion = 0.1 nm times linear dispersion first to, simply the calculations, we can write down the linear dispersion (which has been calculated in (3)) as for grating 1, Linear dispersion = 0.3 times 1/d cos 10 degree = 0.3 (m)/1.667 times 10^-3 (mm) cos 10 degree = 0.1827 (mm/nm) \r\n For crating 2, linear dispersion = 0.3 (m)/8.33 times 10^-4 almostequalto 0.3657 (mm/nm) Therefore, slit width for grating (1) almostequalto 0.1 nm times 0.1827 (mm/nm) = 18.27 mu m and, slit width for grating (2) almostequalto 0.1 nm times 0.3657 (mm/nm) = 36.57 mu m. Out of two gratings, the second one would be the best choice for the spectrometer as grating (2) would provide a larger slit width than grating (1) thus allowing more amount of light to enter the spectrometer...