Find Out if Diffraction Grating Can Resolve Wavelengths 5800 & 5802 Angstroms

[Reshma]

A beam of light is incident normally on a diffraction grating of width 2cm. It is found that at 30 degrees, the n^th order diffraction maximum for \lambda_1 = 5 \times 10^{-5}cm is superimposed on the (n + 1)^th order of \lambda_2 = 4 \times 10^{-5}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 \lambda_3 = 5800 Angstrom units & \lambda_4 = 5802 Angstrom units.

My work:

1]If N ruling occupy a total width W, then slit width d=W/N.

d\sin \theta = n\lambda_1 = (n+1)\lambda_2

{2\over N}{1\over 2} = n\lambda_1 = (n+1)\lambda_2

5000n= 4000(n+1)[/tex] (in Angstrom units).<br /> <br /> So, I got: n = 4; which I substituted in the first equation and I got the total number of rulings <b>N = 0.5 x 10⁴</b><br /> <br /> So, number of rulings per cm is: N/Total width = N/2 = 0.25 x 10⁴<br /> <br /> Is this part correct?<br /> <br /> 2]For this part, I can find the difference between the 2 wavelengths:<br /> <br /> \Delta \lambda = 2 Angstrom units.<br /> How do I determine whether the grating has good resolving power or not?

 

[stunner5000pt]

i think for your scond one you need to find the distance between n=1 nodal lines for each of the wavelengths and compare with the power of resolution of the human eye (look it up)

 

[Reshma]

You mean I have to find the angular separation between the two lines in order to determine whether the resolution is good or not?

 

[Reshma]

Hey someone please help me here...I am unable to get any breakthroughs. Different sources give different resolution power for human eye and I don't think it is needed in this problem. Is there some other way of determining whether the resolution is good or not?

 

[siddharth]

In this problem, the beam of light is inclined at an angle of 30 degrees to the grating.

So,

d\sin \theta = n\lambda_1

won't be right as you will have to factor in the difference in path length because the beam of light is inclined at an angle of 30 degrees to the grating.