Light Diffraction Homework: 2 Wavelengths & Angle of Coincidence

[pat666]

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$$\lambda$$L/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.

 

[rl.bhat]

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.

 

[pat666]

Thanks rl.bhat,
$$\lambda$$₁=6.4*10^-7m
$$\lambda$$₂=4.8*10^-7m
assuming my value for d from my 1st post is correct,
3*6.4*10^-7=4*10-6 sin ($$\theta$$)
$$\theta$$=28.69^o
Does this seem right, mainly concerned about my value for d??

Thanks

 

[rl.bhat]

You are right.

 

[rwisz]

I would first confirm the equations and values used in the attempted solution. The equation for diffraction grating is actually y = nλ/d, where y is the distance between the bright fringes, n is the order of the diffraction, λ is the wavelength, and d is the spacing between the grating lines. The value of d should be 1/2.5 x 10^5 = 4 x 10^-5 m, not 4 x 10^-6 m.

Next, I would use the given information to solve for the two wavelengths and the angle of diffraction at which they coincide. Since the third order of one wavelength coincides with the fourth order of the other, we can set up the following equations:

3λ1 = 4λ2 and y3 = y4

Substituting in the value of d and using the equation for diffraction grating, we get:

3λ1 = 4λ2 and (3λ1)sinθ = (4λ2)sinθ

Simplifying and rearranging, we get:

(3/4)sinθ = (λ2/λ1)

Using the value of Δλ = 160 nm, we can express λ2 in terms of λ1 as:

λ2 = λ1 + 160 x 10^-9

Substituting this into the above equation, we get:

(3/4)sinθ = (λ1 + 160 x 10^-9)/λ1

Solving for θ, we get:

θ = sin^-1(4/7) = 44.4°

Using this angle, we can solve for the two wavelengths:

λ1 = (y3d)/3 = (4.4 x 10^-5)(3)/(3) = 4.4 x 10^-5 m = 440 nm
λ2 = (y4d)/4 = (4.4 x 10^-5)(4)/(3) = 5.9 x 10^-5 m = 590 nm

Therefore, the two wavelengths are 440 nm and 590 nm, and the angle of diffraction at which they coincide is 44.4°. It is worth noting that this solution assumes that the two wavelengths have the same intensity in the diffracted light, which may not be the case in reality. Further experimentation and