Diffraction grating problem involving two wavelengths.

[bfrank]

Homework Statement

A diffraction grating has 4,200 rulings/cm. On a screen 2.00 m from the grating, it is found that for a particular order m, the maximum corresponding to two closely spaced wavelengths of sodium (589.0 nm and 589.6 nm) are separated by 1.54 mm. Determine the value of m.

Homework Equations

dsinθ = mλ (for bright fringes)

The Attempt at a Solution

I spent about 15 minutes working this problem out under the assumption that sinθ ≈ tanθ = y/L. First, I drew a diagram of the corresponding bright fringes, and I made λ₁ correspond to the wavelength that is higher on the screen, while λ₂ is lower on the screen. Since λ₁ has a larger angle in my diagram, I knew that λ₁ should be 589.6 nm for my equations and λ₂ should be 589.0 nm. Also, I knew that y₂ = y₁ - 1.54*10^-3. I came up with two equations:

dsinθ₁ = mλ₁ → dy₁/L = mλ₁
dsinθ₂ = mλ₂ → dy₂/L = mλ₂ → d(y₁ - 1.54*10^-3) = mλ₂

I divided these two equations, went through some algebra, solved for y₁, and plugged it back into the first equation I found to solve for m; I got m = 3. However, I checked the back of the book and it says m = 2. So, I checked over my work again and realized that the small angle approximation I made does not hold because the angles are not small for diffraction gratings (correct me if I'm wrong). Now, I'm back to square one and honestly not sure where to go. I know sinθ = y/hypotenuse, but it seems like this route could get very algebraically sloppy, and I don't want to go down that road before getting some help.

Thank you!

 

[ehild]

dsinθ₁ = mλ₁ → dy₁/L = mλ₁
dsinθ₂ = mλ₂ → dy₂/L = mλ₂ → d(y₁ - 1.54*10^-3) = mλ₂

L is missing from the equation marked wit red.

So you have the equations

dy₁/L = mλ₁*

d(y₁ - 1.54*10^-3)/L = mλ₂ -->

dy₁ /L-1.54*10^-3*d/L=mλ₂**

Plug in mλ₁ for dy₁/L in the second one, and substitute the values for d, L, and for the wavelengths.

ehild

 

[bfrank]

L is missing from the equation marked wit red.

So you have the equations

dy₁/L = mλ₁*

d(y₁ - 1.54*10^-3)/L = mλ₂ -->

dy₁ /L-1.54*10^-3*d/L=mλ₂**

Plug in mλ₁ for dy₁/L in the second one, and substitute the values for d, L, and for the wavelengths.

ehild

I actually had that missing L in my work, that was just a typo on here, sorry about that! I went through the work for the way you suggested (which was much easier than the way I did it, haha), but still got the same answer I got before (m = 3). Any other suggestions? I still think it's because I'm not allowed to make the small angle approximation with this problem, but I'm not sure where to go after knowing that.

Thanks again!

 

[ehild]

Yes, d=0.01/4200=2.381*10^-6. For m=1,
sin(theta) = Lambda/d=589*10^-9/(2.381*10^-6)=0.247 so theta=14.3°, and tan(theta)=0.256.

This angle is rather big to use the small angle approximation, and the other angles, belonging to m=2 or 3 are even bigger.
Let's try the accurate way. y/L =tan(theta).

sin\theta=\frac{tan(\theta)}{\sqrt{1+tan^2(\theta)}},

and substituting tan(θ) with y/L,

sin\theta=\frac{y/L}{\sqrt{1+(y/L)^2}}

Substitute this expression for sin theta in the original equations. You get equations both for y1 and y2. Square them to remove the the root, and solve for the y-s in terms of m. Now try m=1, 2,... and see the difference y1-y2. Take care, calculate with 7 digits or more.

ehild

 

[bfrank]

Yes, d=0.01/4200=2.381*10^-6. For m=1,
sin(theta) = Lambda/d=589*10^-9/(2.381*10^-6)=0.247 so theta=14.3°, and tan(theta)=0.256.

This angle is rather big to use the small angle approximation, and the other angles, belonging to m=2 or 3 are even bigger.
Let's try the accurate way. y/L =tan(theta).

sin\theta=\frac{tan(\theta)}{\sqrt{1+tan^2(\theta)}},

and substituting tan(θ) with y/L,

sin\theta=\frac{y/L}{\sqrt{1+(y/L)^2}}

Substitute this expression for sin theta in the original equations. You get equations both for y1 and y2. Square them to remove the the root, and solve for the y-s in terms of m. Now try m=1, 2,... and see the difference y1-y2. Take care, calculate with 7 digits or more.

ehild

Ah, finally got it! I knew I had to use some trig identity, but I couldn't figure out what. Thank you so much!