- #1
Reshma
- 749
- 6
A beam of light is incident normally on a diffraction grating of width 2cm. It is found that at 30 degrees, the nth order diffraction maximum for [itex]\lambda_1 = 5 \times 10^{-5}[/itex]cm is superimposed on the (n + 1)th order of [itex]\lambda_2 = 4 \times 10^{-5}[/itex]cm.
1]How many lines per cm does the grating have?
2]Find out whether the first order spectrum from such a grating can be used to resolve the wavelengths [itex]\lambda_3 = 5800[/itex] Angstrom units & [itex]\lambda_4 = 5802[/itex] Angstrom units.
My work:
1]If N ruling occupy a total width W, then slit width d=W/N.
[tex]d\sin \theta = n\lambda_1 = (n+1)\lambda_2[/tex]
[tex]{2\over N}{1\over 2} = n\lambda_1 = (n+1)\lambda_2[/tex]
[itex]5000n= 4000(n+1)[/tex] (in Angstrom units).
So, I got: n = 4; which I substituted in the first equation and I got the total number of rulings N = 0.5 x 104
So, number of rulings per cm is: N/Total width = N/2 = 0.25 x 104
Is this part correct?
2]For this part, I can find the difference between the 2 wavelengths:
[itex]\Delta \lambda = 2[/itex] Angstrom units.
How do I determine whether the grating has good resolving power or not?
1]How many lines per cm does the grating have?
2]Find out whether the first order spectrum from such a grating can be used to resolve the wavelengths [itex]\lambda_3 = 5800[/itex] Angstrom units & [itex]\lambda_4 = 5802[/itex] Angstrom units.
My work:
1]If N ruling occupy a total width W, then slit width d=W/N.
[tex]d\sin \theta = n\lambda_1 = (n+1)\lambda_2[/tex]
[tex]{2\over N}{1\over 2} = n\lambda_1 = (n+1)\lambda_2[/tex]
[itex]5000n= 4000(n+1)[/tex] (in Angstrom units).
So, I got: n = 4; which I substituted in the first equation and I got the total number of rulings N = 0.5 x 104
So, number of rulings per cm is: N/Total width = N/2 = 0.25 x 104
Is this part correct?
2]For this part, I can find the difference between the 2 wavelengths:
[itex]\Delta \lambda = 2[/itex] Angstrom units.
How do I determine whether the grating has good resolving power or not?