Visible light is incident perpendicularly on a diffraction grating of 200 rulings/mm. What are the (a) longest, (b) second longest, and (c) third longest wavelengths that can be associated with an intensity maximum at theta=30 degrees?

Question

Visible light is incident perpendicularly on a diffraction grating of 200 rulings/mm. What are the (a) longest, (b) second longest, and (c) third longest wavelengths that can be associated with an intensity maximum at theta=30 degrees?

Answer 1

d•sinθ=k•λ

N₀=1/d =>
λ=sinθ/k•N₀
sin30°=0.5, N₀=2•10⁵ m⁻¹
k=1:
λ=sinθ/k•N₀=sin30°/1•2•10⁵=2.5•10⁻⁶ m
invisible - Infra-red
k=2
λ=sinθ/k•N₀=sin30°/2•2•10⁵=1.25•10⁻⁶ m
invisible - Infra-red

k=3
λ=sinθ/k•N₀=sin30°/3•2•10⁵=8.33•10⁻⁷ m
invisible - Infra-red
k=4 ….!!!!!
λ=sinθ/k•N₀=sin30°/4•2•10⁵=6.25•10⁻⁷ m
visible - red
k=5…. !!!!
λ=sinθ/k•N₀=sin30°/5•2•10⁵=5.0•10⁻⁷ m
visible - green
k=6 ….!!!!
λ=sinθ/k•N₀=sin30°/6•2•10⁵=4.17•10⁻⁷ m
visible - violet
k=7
λ=sinθ/k•N₀=sin30°/7•2•10⁵=3.57•10⁻⁷ m
invisible- ultra-violet

Answer 2

To find the longest, second longest, and third longest wavelengths associated with an intensity maximum at a given angle, we can use the formula for the grating equation:

nλ = d(sinθ ± sinφ)

where:
n = order of the maximum (1 for the first, 2 for the second, and so on)
λ = wavelength of light
d = spacing between adjacent rulings on the grating
θ = angle of incidence (in this case, 30 degrees)
φ = angle of diffraction

In this case, the grating has 200 rulings per millimeter, which means the spacing between adjacent rulings (d) is given by:

d = 1 / (200 x 10^6) mm

Now let's find the longest, second longest, and third longest wavelengths using n = 1, 2, and 3 respectively:

(a) Longest wavelength:
For n = 1, the equation becomes:
λ₁ = d(sinθ ± sinφ₁)

We need to find the value of φ₁. Since the light is incident perpendicularly, φ₁ = θ, so we have:
λ₁ = d(2sinθ)

Substituting the given values:
λ₁ = (1 / (200 x 10^6) mm)(2sin(30°))

Now we can convert the result to a more common unit, such as nanometers (nm).

(b) Second longest wavelength:
For n = 2, the equation becomes:
λ₂ = d(sinθ ± sinφ₂)

We need to find the value of φ₂. By comparing orders 1 and 2, we can determine that:
sinφ₂ = 2sin(30°) - sin(60°)

(c) Third longest wavelength:
For n = 3, we can use a similar process to find λ₃.

Now we can calculate λ₂ and λ₃ using the same approach as for λ₁. Remember to convert the results to a common unit.

By following these steps, you should be able to find the longest, second longest, and third longest wavelengths associated with an intensity maximum at θ = 30 degrees.