Diffraction Physics Example 4

Follow the full solution, then compare it with the other examples linked below.

Example 4

hard
A diffraction grating has 500 lines/mm500 \text{ lines/mm}. Light of wavelength 650 nm650 \text{ nm} passes through it. At what angle does the second-order maximum appear?

Solution

  1. 1
    Slit spacing: d=1500 mm=2×10−6 md = \frac{1}{500} \text{ mm} = 2 \times 10^{-6} \text{ m}.
  2. 2
    Grating equation for maxima: dsin⁡θ=mλd\sin\theta = m\lambda. For m=2m = 2:
  3. 3
    sin⁡θ=mλd=2×650×10−92×10−6=1.3×10−62×10−6=0.65  ⟹  θ=sin⁡−1(0.65)≈40.5°\sin\theta = \frac{m\lambda}{d} = \frac{2 \times 650 \times 10^{-9}}{2 \times 10^{-6}} = \frac{1.3 \times 10^{-6}}{2 \times 10^{-6}} = 0.65 \implies \theta = \sin^{-1}(0.65) \approx 40.5°

Answer

θ≈40.5°\theta \approx 40.5°
A diffraction grating produces sharp interference maxima at specific angles. More lines per mm means wider spacing between orders. The grating equation dsin⁡θ=mλd\sin\theta = m\lambda determines where bright fringes appear.

About Diffraction

The spreading of a wave as it passes through a gap or around the edge of an obstacle.

Learn more about Diffraction →

More Diffraction Examples

Example 1 easy

Sound waves ([formula]) pass through a doorway [formula] wide. Do the sound waves diffract significa

Example 2 medium

Light of wavelength [formula] passes through a single slit of width [formula]. At what angle does th

Example 3 medium

Visible light ([formula]) and radio waves ([formula]) both encounter a building [formula] wide. Whic

← All Diffraction Examples Diffraction Concept