Multiple slits IB DP Physics Study Notes - 2025 Syllabus
Multiple slits IB DP Physics Study Notes
Multiple slits IB DP Physics Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on IB Physics syllabus with Students should understand
Young’s double-slit interference as given by $s =\frac{ λD}{d}$ where s is the separation of fringes, d is the
separation of the slits, and D is the distance from the slits to the screen
Standard level and higher level: 5 hours
Additional higher level: 6 hours
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
Double-slit interference
- Double-slit interference is a phenomenon that occurs when coherent light passes through two closely spaced slits and produces an interference pattern on a screen.
- This experiment, first performed by Thomas Young in 1801, provided strong evidence for the wave nature of light.
- The result is a pattern of alternating bright and dark fringes due to constructive and destructive interference of light waves.
- Constructive interference (bright fringes) occurs when the path difference between waves from the two slits is an integer multiple of the wavelength: \( \Delta L = n\lambda \)
- Destructive interference (dark fringes) occurs when the path difference is a half-integer multiple of the wavelength: \( \Delta L = (n + \frac{1}{2})\lambda \)
Key equation:
- \( s \) = fringe separation (distance between adjacent bright or dark fringes)
- \( \lambda \) = wavelength of the light
- \( D \) = distance from the double-slit to the screen
- \( d \) = distance between the two slits
Example:
In a double-slit experiment, the slit separation is \( d = 0.25 \, \text{mm} \), the distance to the screen is \( D = 2.0 \, \text{m} \), and the light used has a wavelength of \( \lambda = 600 \, \text{nm} \). Calculate the distance between adjacent bright fringes.
▶️ Answer/Explanation
Step 1: Convert all quantities to SI units
\( \lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \)
\( d = 0.25 \, \text{mm} = 0.25 \times 10^{-3} \, \text{m} \)
\( D = 2.0 \, \text{m} \)
Step 2: Use the formula
\( s = \frac{\lambda D}{d} = \frac{600 \times 10^{-9} \times 2.0}{0.25 \times 10^{-3}} \)
Step 3: Calculate
\( s = \frac{1.2 \times 10^{-6}}{0.25 \times 10^{-3}} = 4.8 \times 10^{-3} \, \text{m} = \boxed{4.8 \, \text{mm}} \)
Final Answer:
The distance between adjacent bright fringes is \( \boxed{4.8 \, \text{mm}} \).
IB Physics Multiple slits Exam Style Worked Out Questions
Question
Monochromatic light is incident on a single slit to form a diffraction pattern on a screen. The width of the single slit is then halved.
What are the change in the width of the central maximum and the change in the maximum intensity of the pattern?
▶️Answer/Explanation
Ans:C
Question
In two different experiments, white light is passed through a single slit and then is either refracted through a prism or diffracted with a diffraction grating. The prism produces a band of colours from $\mathrm{M}$ to $\mathrm{N}$. The diffraction grating produces a first order spectrum $\mathrm{P}$ to $\mathrm{Q}$.
What are the colours observed at $\mathrm{M}$ and $\mathrm{P}$ ?
▶️Answer/Explanation
Ans:A