Diffraction grating, distances between maxima

[TheKShaugh]

Homework Statement

A grating has a line density of 1010 cm−1, and a screen perpendicular to the ray that makes the central peak of the diffraction pattern is 2.5 m from the grating. If light of two wavelengths, 590 nm and 680 nm, passes through the grating, what is the separation on the (flat) screen between the second-order maxima for the two wavelengths?

Homework Equations

dsin\theta = m \lambda

The Attempt at a Solution

I thought I would find the angular distance to the second order maxima covered by each wavelength:

\theta = arcsin(\frac{2 \lambda}{d}) where d is \frac{1}{1010}\times 10^{-2}

Then I would take the difference, and plug it into the equation sin\theta = \frac{y}{L} and solve for y. This gives me the wrong answer but I don't see why this wouldn't work. Can anyone see my mistake?

Thanks!

 

[ehild]

Homework Statement

A grating has a line density of 1010 cm−1, and a screen perpendicular to the ray that makes the central peak of the diffraction pattern is 2.5 m from the grating. If light of two wavelengths, 590 nm and 680 nm, passes through the grating, what is the separation on the (flat) screen between the second-order maxima for the two wavelengths?

Homework Equations

dsin\theta = m \lambda

The Attempt at a Solution

I thought I would find the angular distance to the second order maxima covered by each wavelength:

\theta = arcsin(\frac{2 \lambda}{d}) where d is \frac{1}{1010}\times 10^{-2}

Then I would take the difference, and plug it into the equation sin\theta = \frac{y}{L} and solve for y.

Take the difference of what?
You have two wavelengths, two angles and two y positions. You need the separation between the positions, that is, the difference of the y-s.

 

[TheKShaugh]

Take the difference of what?
You have two wavelengths, two angles and two y positions. You need the separation between the positions, that is, the difference of the y-s.

What I meant was that I would solve for the angular spread of the second order maxima for one wavelength, then the other, and the take the difference of those two values. What I get should be the angular distance between the two maxima. I could then use the equation I gave in my OP and solve. This gives the same answer that I got: 2.5sin(7.896)-2.5sin(6.845) The idea is that if I calculate the angle to one maxima, I should be able to tell what vertical distance it covers. Then I can do the same for the other maxima, and take the difference between those two distances as the distance from one maxima to another.

 

[ehild]

It is a good method, what result did you get? In principle, the distance of the maximum from the centre is L tan(θ). If θ is really small, the sine and the tangent are nearly equal. But the angles you got are not small enough.