AP Physics 2- 14.8 Double-Slit Interference and Diffraction Gratings- Study Notes- New Syllabus
AP Physics 2- 14.8 Double-Slit Interference and Diffraction Gratings – Study Notes
AP Physics 2- 14.8 Double-Slit Interference and Diffraction Gratings – Study Notes – per latest Syllabus.
Key Concepts:
- Double-Slit Interference
- Diffraction Gratings
Double-Slit Interference
Double-slit interference occurs when coherent light passes through two narrow slits separated by a distance comparable to the wavelength of light. The overlapping waves from the two slits interfere, producing alternating bright and dark fringes on a screen.
Interference patterns produced by light interacting with a double slit indicate that light has wave properties. The source of this discovery was Young’s double-slit experiment.
Visual representations of double-slit diffraction patterns are useful in determining the physical properties of the slits and the interacting waves.
Path Difference (\(\Delta D\)):
The path difference between the light from the two slits is the extra distance traveled by one wave relative to the other:
$ \mathrm{\Delta D = d \sin \theta} $
- \(\mathrm{d}\) = separation between the two slits
- \(\theta\) = angle of the fringe on the screen relative to the central maximum
Condition for Interference:
Constructive Interference (Bright Fringe): Occurs when the path difference is an integer multiple of the wavelength:
$ \mathrm{\Delta D = d \sin \theta = m \lambda}, \quad m = 0, 1, 2, \dots $
Destructive Interference (Dark Fringe): Occurs when the path difference is an odd multiple of half the wavelength:
$ \mathrm{\Delta D = d \sin \theta = (m + \frac{1}{2}) \lambda}, \quad m = 0, 1, 2, \dots $
Small-Angle Approximation:
When \(\theta\) is small (\(\theta \lesssim 10^\circ\)), we can approximate:
$ \mathrm{\sin \theta \approx \tan \theta \approx \theta \text{ (in radians)}} $
Then the distance from the central maximum to the m-th fringe on a screen at distance \(\mathrm{L}\) is:
$ \mathrm{y_m \approx L \tan \theta_m \approx L \frac{\Delta D}{d} = L \frac{m \lambda}{d}} $
Fringe Width:
$ \mathrm{w = y_{m+1} – y_m = \frac{L \lambda}{d}} $
Example :
Light of wavelength \(\mathrm{500 \, nm}\) passes through two slits separated by \(\mathrm{0.2 \, mm}\). The screen is \(\mathrm{2 \, m}\) away. Find the distance of the 3rd bright fringe from the central maximum.
▶️ Answer/Explanation
Step 1: Path difference for 3rd bright fringe: \(\mathrm{\Delta D = m \lambda = 3 \times 500 \, nm = 1500 \, nm}\)
Step 2: Use small-angle approximation: \(\mathrm{y_m = L \frac{\Delta D}{\lambda}}?\) Actually, directly: \(\mathrm{y_m \approx L \frac{m \lambda}{d}}\)
Step 3: Substitute values: \(\mathrm{y_3 = 2 \times \frac{3 \times 500 \times 10^{-9}}{0.2 \times 10^{-3}} = 0.015 \, m = 1.5 \, cm}\)
Final Answer: \(\mathrm{y_3 = 1.5 \, cm}\)
Example:
A double-slit experiment uses light of wavelength \(\mathrm{600 \, nm}\) and slit separation \(\mathrm{0.25 \, mm}\). The screen is \(\mathrm{1.5 \, m}\) away. Calculate the fringe width.
▶️ Answer/Explanation
Step 1: Fringe width formula: \(\mathrm{w = \frac{L \lambda}{d}}\)
Step 2: Substitute values: \(\mathrm{w = \frac{1.5 \times 600 \times 10^{-9}}{0.25 \times 10^{-3}} = 3.6 \times 10^{-3} \, m = 3.6 \, mm}\)
Final Answer: Fringe width = \(\mathrm{3.6 \, mm}\)
Diffraction Gratings
A diffraction grating consists of a large number of equally spaced parallel slits. When light passes through or reflects from the grating, it undergoes diffraction, producing bright and dark fringes due to interference.
Formation of Maxima:
The condition for principal maxima is given by:
$ \mathrm{d \sin \theta = m \lambda}, \quad m = 0, 1, 2, \dots $
- \(\mathrm{d}\) = distance between adjacent slits (grating spacing)
- \(\theta\) = angle at which the m-th order maximum occurs
- \(\mathrm{\lambda}\) = wavelength of the light
- \(m\) = order of the maximum
White Light Diffraction:
- When white light is incident on a diffraction grating, the central maximum (m = 0) appears white because all wavelengths overlap.
- Higher-order maxima (\(m \geq 1\)) separate the white light into its constituent colors due to differences in wavelength.
- Red light (\(\lambda \approx 700 \, nm\)) appears farthest from the central maximum, while violet (\(\lambda \approx 400 \, nm\)) appears closest.
Angular Dispersion:
The angular separation of different wavelengths is greater for higher orders (larger \(m\)) and smaller grating spacing \(d\).
Example :
A diffraction grating has 5000 slits per cm. Light of wavelength \(\mathrm{600 \, nm}\) is incident on the grating. Find the angle of the first-order maximum (\(m = 1\)).
▶️ Answer/Explanation
Step 1: Grating spacing: \(\mathrm{d = \frac{1}{5000 \, cm^{-1}} = \frac{1}{5 \times 10^5 \, m^{-1}} = 2.0 \times 10^{-6} \, m}\)
Step 2: Use grating equation: \(\mathrm{d \sin \theta = m \lambda}\)
$\mathrm{\sin \theta = \frac{m \lambda}{d} = \frac{1 \times 600 \times 10^{-9}}{2.0 \times 10^{-6}} = 0.3}$
Step 3: Angle of first-order maximum: \(\mathrm{\theta = \arcsin(0.3) \approx 17.5^\circ}\)