# Diffraction grating problem

1. Jan 30, 2006

### twiztidmxcn

Hey

Here is the problem, I think I may be stuck but I'll provide all the information I have so far.

White light (400-700nm) is incident on a 600 line/mm diffraction grating. What is the width of the first order rainbow on a screen 2.0m behind the grating?

What I figured was to use the equation d*sin(theta) = m * wavelength

as well as the equation y = L*tan(theta), with L = 2.0m, m = 1 (first order rainbow),

Basically, I have to find theta, then find y, which ends up being the width of my projected rainbow. I am, however, totally stuck at this point. I'm not even sure if my assumption that m = 1 is right, or if I am on the right path.

Any help in the right direction would be awesome.

Thank you
twiztidmxcn

2. Jan 30, 2006

### finchie_88

This is probably not the best way, but the way that I would do that is to find theta when the wavelength is 400 nm, and then do the same to find the angle when the wavelength is 700 nm, then use a little trig and the difference between the angles to calculate the width of the spectrum.

calculate the angles by using
$$n\lambda = dsin\theta$$ (n = 1, d = distance between the slits).

Last edited by a moderator: Jan 30, 2006
3. Jan 30, 2006

### twiztidmxcn

theta @ 400nm = 13.9 degrees
theta @ 700nm = 24.8 degrees

using the idea of finding the angle difference (10.9 degrees), then using trig we find that tan(theta) = y / x, y = tan (theta) * x, where x = 2m and theta is the angle difference that i calculated?

thus leaving me with something like y = 2m * tan (10.9) = .385m?

Last edited: Jan 30, 2006
4. Jan 30, 2006

### Staff: Mentor

The difference between the angles doesn't help you here because $\tan{\theta_2} - \tan{\theta_1}$ does not equal $\tan (\theta_2 - \theta_1)$.

5. Jan 30, 2006

### twiztidmxcn

I think I get what you mean

It makes more sense with that explanation, and helps me to reason that since x*tan(theta) = y, y2 - y1 = x(tan(theta2)-tan(theta1)).

What you were saying is to use the difference of lengths rather than angle because... well of what you said.