Diffraction grating question

[Nylex]

How many lines per millimetre should the grating have if for the first order, observed at a wavelength λ = 500 nm, the reflected beam is observed at 10 deg from the normal, the incident angle being 60 deg?

I've done the question and just want to know if I've done anything wrong, as my answer seems too big.

mλ = d(sin i + sin θ)

d = mλ/(sin i + sin θ)

d = (1 x 500 x 10^-9)/(sin 60 + sin 10)

d = 4.809 x 10^-7 m (distance between slits)

No. of lines per m = 1/(4.809 x 10^-7)

=> No. of lines per mm = 1/(4.809 x 10^-7 x 10^-3) = 2.079 x 10^9

Thanks.

[dextercioby]

How many lines per millimetre should the grating have if for the first order, observed at a wavelength λ = 500 nm, the reflected beam is observed at 10 deg from the normal, the incident angle being 60 deg?

I've done the question and just want to know if I've done anything wrong, as my answer seems too big.

mλ = d(sin i + sin θ)

d = mλ/(sin i + sin θ)

d = (1 x 500 x 10^-9)/(sin 60 + sin 10)

d = 4.809 x 10^-7 m (distance between slits)

No. of lines per m = 1/(4.809 x 10^-7)

=> No. of lines per mm = 1/(4.809 x 10^-7 x 10^-3) = 2.079 x 10^9

Thanks.

I don't mean to be rude,but your problems stretch way beyound the realms of statistical mechanics.It' s like a computer virus that now is affecting key files of your memory,making the CPU run slower and give erroneous results.
No.of lines per m=2.079 x 10^6 =>No of lines per mm=2.079 x 10^3.
You can have more lines in one mm than in one m,as long as the density of lines is constant across the grating,right???

[Nylex]

And you spelt beyond wrong. All I did was convert metres to millimetres on the bottom.

[dextercioby]

And you spelt beyond wrong. All I did was convert metres to millimetres on the bottom.

I'm sorry for having offended u in any way,but you're wrong:i spelt "beyond" (sic) in a wrong way.Silly me... :rofl: And the conversion was wrong.That's why it gave u 1000000 times more lines/mm than it should have.

Daniel.