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Physics 214 Problem 3 Week 3 Diffraction Grating
A diffraction grating consisting of N slits produces a second-order maximum for red light (λ = 630 nm) at θ = 35°.
a) Determine the spacing between the slits of the grating.
sin θ = m λ /d , so d = 2 ×××× 0.63 (^) μμμμ m /sin(35) = 2.2 μμμμ m = 2.2 ×××× 10 -6^ m.
b) Assuming a total area of 5 x 5 cm^2 as shown, how many slits (lines) are there in the grating?
N = 0.05 m / 2.2 ×××× 10 -6^ m = 22700 slits.
c) Assuming that the grating is uniformly illuminated over half of its width, how wide an angular width, ∆θ, does the zeroth-order maximum have (see figure)? To calculate this, you first need to find the phase angle φ=φο (between the phasors from adjacent slits) corresponding to the first minimum as shown below. To see what the formula is for several slits, there are two approaches (you can choose which to use): Phasors: First draw a phasor diagram for the case of 8 slits. Here is some help:
The 8-slit phasor diagram for the first zero is:
That is, the phase difference between adjacent slits is φ o = 2π/8. In the N -slit case, φ = 2π/ N in order to form an N sided polygon.
5 cm
5 cm
Face view of screen the grating
θ
grating
Solution
Draw a phasor diagram for this angle, θο. What is φο for one phasor here?
θ 1
8d
λ
Algebraic: solve the N-slit intensity formula for the first zero in φ, then solve for the
corresponding value of θ:
2 1
sin( / 2) N sin( / 2)
N
I I
φ φ
.
The numerator above will go to zero when N φ o /2 = π, or φ o = 2π/N, the same result as above.
Now relate φο to θο. Don’t forget that we want the zero-to-zero width, i.e., ∆θ corresponds to 2 θo. Also, don’t forget that only half of the grating is illuminated.
Since only half the grating is illuminated, the relevant N = 11350 slits. From φ = 2π (δ/λ) = 2π (d sin θ)/λ, we have: sin θo = (φo/2π)(λ/d) = (1/Ν)(λ/d). Since θo is << 1, we can use the small-angle approximation:
θo ≈ (λ/Nd) = 0.633 μm /(11350 x 2.2 μm) = 2.5 x 10-5.
The angular width is then: ∆θ = 2 θo = 5 x 10-5^ radians = (1.45 x 10-3)°
d) Can this grating be used to distinguish two lasers, one at 630 nm, the other at 631 nm?
To distinguish different wavelengths, we need Rayleigh’s criteria: ∆ λ / λ > 1/ N. Here we certainly have 1 / 630 > 2 / 22700, so the answer is ‘yes’. If we illuminate half the grating, we can easily resolve these wavelengths in second order (and even in first order!).
e) How many principal orders of 630-nm light can be seen with this grating? (Only count on one side, and do not count the central [0th-order] peak.)
The diffraction angle can never be more than 90˚:
From sinθ = mλ/d, it must be that mλ ≤ d, or m ≤ d/λ = 2.2 μm/0.63 μm = 3.5 m = 3.
