# Homework Help: Diffraction grating

1. Mar 14, 2006

### Reshma

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 $\lambda_1 = 5 \times 10^{-5}$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.

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

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

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

2. Mar 14, 2006

### 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)

3. Mar 16, 2006

### 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?

4. Mar 16, 2006

### 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?

5. Mar 18, 2006

### 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.

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