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

Monochromatic light of wavelength 600 nm600\text{ nm} is incident normally on a diffraction grating having 500 lines per mm500\text{ lines per mm}. What is the angle of diffraction, in degrees, for the first-order principal maximum?

Answer: 17.5 degrees

Answer

The angle of diffraction for the first-order principal maximum is 17.5∘17.5^\circ.
Using the grating equation dsin⁡θ=nλd \sin \theta = n \lambda, the slit separation is d=10−3 m500=2.00×10−6 md = \frac{10^{-3}\text{ m}}{500} = 2.00 \times 10^{-6}\text{ m}. For n=1n = 1 and λ=6.00×10−7 m\lambda = 6.00 \times 10^{-7}\text{ m}, we get sin⁡θ=6.00×10−72.00×10−6=0.30\sin \theta = \frac{6.00 \times 10^{-7}}{2.00 \times 10^{-6}} = 0.30. Taking arcsin⁡(0.30)\arcsin(0.30) gives approximately 17.5∘17.5^\circ.

Step-by-Step Solution

1
Calculate the grating element (slit spacing) dd
d=2.00×10−6 md = 2.00 \times 10^{-6}\text{ m}
Grating spacing dd is the reciprocal of the line density N=500 lines/mm=500,000 lines/mN = 500\text{ lines/mm} = 500,000\text{ lines/m}.
2
Apply the diffraction grating equation dsin⁡θ=nλd \sin \theta = n \lambda
sin⁡θ=0.30\sin \theta = 0.30
For the first-order maximum (n=1n = 1), sin⁡θ=1×600×10−9 m2.00×10−6 m=0.30\sin \theta = \frac{1 \times 600 \times 10^{-9}\text{ m}}{2.00 \times 10^{-6}\text{ m}} = 0.30.
3
Find the angle θ\theta by taking the inverse sine
θ=17.5∘\theta = 17.5^\circ
arcsin⁡(0.30)≈17.46∘\arcsin(0.30) \approx 17.46^\circ, which rounds to 17.5∘17.5^\circ.

Key Concept

Diffraction Grating Equation for Principal Maxima
Estimated Time:1m 30s
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Topics

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

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