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Difficulty: MediumWave Phenomena: Reflection, Refraction, Interference, Diffraction, and Polarization

A beam of monochromatic light of wavelength 6.25×10−7 m6.25 \times 10^{-7}\text{ m} is incident normally on a plane diffraction grating having 400 lines/mm400\text{ lines/mm}. What is the angle of diffraction for the second-order principal maximum?

  1. A
    14.5∘14.5^\circ
  2. 30.0∘30.0^\circAnswer
  3. C
    45.0∘45.0^\circ
  4. D
    60.0∘60.0^\circ

Answer

The angle of diffraction for the second-order principal maximum is 30.0∘30.0^\circ.
The grating spacing is d=1400×103 m−1=2.50×10−6 md = \frac{1}{400\times 10^3\text{ m}^{-1}} = 2.50 \times 10^{-6}\text{ m}. Using dsin⁡θ=nλd \sin\theta = n\lambda for n=2n=2 and λ=6.25×10−7 m\lambda = 6.25 \times 10^{-7}\text{ m} gives sin⁡θ=2×6.25×10−72.50×10−6=0.50\sin\theta = \frac{2 \times 6.25 \times 10^{-7}}{2.50 \times 10^{-6}} = 0.50, corresponding to θ=30.0∘\theta = 30.0^\circ.

Step-by-Step Solution

1
Calculate the grating spacing (dd).
d=1 mm400=2.50×10−3 mm=2.50×10−6 md = \frac{1\text{ mm}}{400} = 2.50 \times 10^{-3}\text{ mm} = 2.50 \times 10^{-6}\text{ m}.
Grating spacing dd is the reciprocal of the line density NN per unit length.
2
Apply the diffraction grating equation for the second order (n=2n=2).
dsin⁡θ=nλ  ⟹  (2.50×10−6)sin⁡θ=2×(6.25×10−7)=1.25×10−6d \sin\theta = n\lambda \implies (2.50 \times 10^{-6}) \sin\theta = 2 \times (6.25 \times 10^{-7}) = 1.25 \times 10^{-6}.
The principal maximum condition relates slit spacing, diffraction angle, spectral order, and wavelength.
3
Solve for the diffraction angle θ\theta.
sin⁡θ=1.25×10−62.50×10−6=0.50  ⟹  θ=arcsin⁡(0.50)=30.0∘\sin\theta = \frac{1.25 \times 10^{-6}}{2.50 \times 10^{-6}} = 0.50 \implies \theta = \arcsin(0.50) = 30.0^\circ.
Taking the inverse sine of 0.500.50 yields the exact angle of diffraction.

Key Concept

Diffraction Grating Equation (dsin⁡θ=nλd \sin\theta = n\lambda)
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Topics

Wave Phenomena: Reflection, Refraction, Interference, Diffraction, and Polarization

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