A diffraction grating having 180 lines/mm is illuminated with a light signal containing
two wavelengths, λ1 = 400 nm and λ2 = 500 nm. The signal is incident perpendicularly
on the grating. What is the smallest angle at which two of the resulting maxima are superimposed?
λ1=400nm=destructive?: 2t=1/2(λ/n)+(m)(λ/n)=?
λ2=500nm= constructive?: 2t=m(λ/n)=?
I think this is what im supposed to do but again I could be wrong.
To find the smallest angle at which two of the resulting maxima are superimposed, we can use the formula for the maximum order (m) of a diffraction grating:
m(λ/n) = d * sinθ
Where:
- m is the order of the maxima, which can be positive or negative integer values (in this case, we are interested in consecutive orders)
- λ is the wavelength of the light signal
- n is the index of refraction of the medium the light is passing through (which can be assumed to be 1 for air or vacuum)
- d is the grating spacing, which is the inverse of the number of lines per unit length (d = 1/lines per mm)
For the 400 nm light signal (λ1 = 400 nm):
We want to find the angle where this wavelength is destructive interference (superimposed). This means the path difference between the two slits of the grating, considering the mth order (m can be positive or negative), should be half the wavelength:
2t = (1/2)(λ1/n) + m(λ1/n)
2t = (1/2)(400 nm/1) + m(400 nm/1)
2t = 200 nm + 400m nm
For the 500 nm light signal (λ2 = 500 nm):
We want to find the angle where this wavelength is constructive interference (not superimposed). This means the path difference between the two slits of the grating, considering the mth order (m can be positive or negative), should be equal to the wavelength:
2t = m(λ2/n)
2t = m(500 nm/1)
2t = 500m nm
Now, we set these two equations equal to each other to find the smallest angle at which the two maxima are superimposed (destructive interference):
200 nm + 400m nm = 500m nm
Rearrange the equation to solve for m:
200 nm = 100m nm
m = 2
Now, substitute the value of m into the formula for finding the angle of incidence:
2t = 500m nm
2t = 500(2) nm
t = 500 nm/2
t = 250 nm
The angle θ can be obtained by using the formula: sinθ = t/d
sinθ = (250 nm) / (1/180 lines/mm)
sinθ = (250 nm) / (0.001 mm)
sinθ = 0.25
To find the smallest angle, we take the inverse sine of both sides:
θ = sin^(-1)(0.25)
θ ≈ 14.48 degrees
Therefore, the smallest angle at which two of the resulting maxima are superimposed is approximately 14.48 degrees.