Light Diffraction Homework: 2 Wavelengths & Angle of Coincidence

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The problem involves light with two wavelengths differing by 160 nm passing through a diffraction grating with 250,000 lines per meter. The third-order maximum of one wavelength coincides with the fourth-order maximum of the other. The calculations confirm that the wavelengths are approximately 640 nm and 480 nm, with the angle of diffraction at about 28.69 degrees. The value for the grating spacing (d) was verified to be correct at 4 x 10^-6 m. The solution effectively demonstrates the relationship between the wavelengths and their respective diffraction orders.
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Homework Statement



12 Light consisting of two wavelengths which differ by 160 nm passes through a diffraction grating with 2.50 x 105 lines per metre. In the diffracted light, the third order of one wavelength coincides with the fourth of the other. What are the two wavelengths and at what angle of diffraction does this coincidence occur?

Homework Equations



y=n\lambdaL/d I think is what I should be looking at??

The Attempt at a Solution


\Delta\lambda=160*10^-9
d=1/2.5*10^5 = 4*10-6 not sure if this part is right
If my d is right I think that what I should do next is
y3=3*\lambda*L/4*10^-6
y4=4*\lambda*L/4*10^-6
these are both bright fringes but maybe one should be dark?
That is as far as I get at the moment!

Thanks for any help.
 
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Formula for diffraction grating is

d*sin(θ) = mλ

at the point of coincidence θ is same for both the wavelengths. So

m1λ1 = m2λ2

3*λ1 = 4(λ1- 160)

Solve for λ1. And then proceed to find the other results.
 
Thanks rl.bhat,
\lambda1=6.4*10-7m
\lambda2=4.8*10-7m
assuming my value for d from my 1st post is correct,
3*6.4*10-7=4*10-6 sin (\theta)
\theta=28.69o
Does this seem right, mainly concerned about my value for d??

Thanks
 
You are right.
 
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