00:01
We are given a diffraction grating with line density 5 ,000 per centimeter and the width of the diffraction grading itself is 3 .5 centimeters.
00:11
So the total number of lines on the diffraction grating is the density times the length, which is 5 ,000 times 3 .5, which is 17 ,500 lines.
00:32
So that now we know that chromatic resolving power is lambda by delta lambda.
00:39
And for that fraction grading that is nm where n is the number of slits and m is the order at which we are resolving so for order one it's simply n which is 17 ,500 now that we have this for part b we want to see if we can resolve the sodium doublet line sodium doublet line has the wavelength 589 nanometers and 589 nanometers so lambda is any of these so so delta lambda is the difference which is 0 .59 nanometers.
01:18
So now the chromatic resolving power is lambda by delta lambda, which is now because the difference is so small, for lambda we can consider either of these.
01:27
So that's 589 divided by delta lambda is 0 .59, which is almost 1 ,000.
01:35
So because our resolving power is larger than this, we can successfully resolve the sodium now part c has multiple questions.
01:47
The first thing to note is that we are operating at order 2 now...