Derivation of Intensity Formula and Its Maximum and Minimum Values in YDSE
Intensity In Young's Double Slit Experiment is a must-know concept for every JEE Main aspirant because it lies at the heart of understanding interference patterns in wave optics. The analysis of intensity allows you to explain why alternate bright and dark bands form on the screen and to solve a wide range of numerical problems quickly.
Accurately deriving the intensity formula, recognizing the factors affecting maximum and minimum values, and applying these ideas in Physics questions gives you a real edge in exams like JEE.
At the core of the Young's double slit experiment derivation lies the principle that two coherent light sources create points of constructive and destructive interference on a screen, leading to a pattern of fringes. The intensity at each screen point reflects the superposition of light waves, an idea rooted in the superposition principle of waves. Mastering this, along with the key equations, helps you crack both theory and numerical questions easily.
Definition and Overview of Intensity In Young's Double Slit Experiment
In intensity in Young's double slit experiment, intensity refers to the power received per unit area at a specific point on the screen. This arises due to the combination of two light waves emanating from the slits, whose amplitudes superpose and interfere. When the waves combine constructively, intensity peaks, creating bright fringes. Destructive interference leads to minima and dark fringes.
This intensity distribution is critical because it distinguishes interference of light. from simple overlapping and is foundational in wave optics. The experiment exemplifies how coherent sources are essential for observing clear interference patterns.
Experimental Setup and Intensity Measurement in YDSE
The basic YDSE setup features:
- A monochromatic light source illuminating two narrow slits.
- Slits S1 and S2 placed at a small separation d.
- A screen positioned at a distance D from the slits (with D ≫ d).
- Measurement of intensity at various points on the screen perpendicular to the slits.
At any position on the screen, the intensity depends on the path difference between waves from S1 and S2. This forms a series of alternating bright and dark bands, known as fringes. For solid revision on setups, refer to the optics page and the coherence and coherent sources guide.
Mathematical Derivation of Intensity In Young's Double Slit Experiment
The derivation of the intensity distribution in Young's double slit experiment starts from the superposition of two waves with amplitudes a1 and a2. For equal intensity and phase difference ϕ:
- Write the electric field due to each slit: E1, E2.
- Use the principle of superposition: E = E1 + E2.
- Express resultant intensity: I ∝ (E1 + E2)².
- For equal amplitude a and maximum intensity I0 = 4a², intensity at phase difference ϕ is:
| Formula | Meaning |
|---|---|
| I = I0 cos²(ϕ/2) | I = intensity at a point, I0 = maximum intensity, ϕ = phase difference |
| ϕ = (2π/λ)(d sinθ) | ϕ relates path difference to position on the screen |
The key here is that the resultant intensity varies as the square of the cosine of half the phase difference, directly producing the alternating maxima and minima observed.
Intensity Distribution and Intensity Graph in YDSE
The intensity distribution in Young's double slit experiment forms a sinusoidal pattern along the screen. As you move away from the central bright fringe (θ = 0), intensity varies between maxima and minima.
- Central maximum (ϕ = 0): I = I0 (brightest point)
- Next maxima at ϕ = 2nπ: I = I0
- Minima (dark fringes) at ϕ = (2n+1)π: I = 0
- Fringes are equally spaced, determined by d, D, and λ
- Relative intensity falls off symmetrically from center
A graph of intensity versus position is a series of equally spaced peaks (maxima) and troughs (minima). This reflects perfect wave superposition between the two slits. For more on pattern nuances, see diffraction of light for real-world edge cases.
Maximum, Minimum, and Average Intensity: Key YDSE Formulas and Example
Maximum intensity in YDSE occurs at bright fringes (constructive interference) where ϕ is an even multiple of π. For equal intensity from both slits:
- Maximum intensity (Imax): Imax = 4I0 when both amplitudes are equal
- Minimum intensity (Imin): Imin = 0 (dark fringes at odd multiples of π phase difference)
- Average intensity: Iavg = (Imax + Imin)/2
| Quantity | Formula | Key Condition |
|---|---|---|
| Imax | 4I0 | Constructive, equal amplitudes |
| Imin | 0 | Destructive, equal amplitudes |
| I at any point | I = I0 cos²(ϕ/2) | Phase difference ϕ |
For example, if I0 = 1 unit at a point where ϕ = 60°, then I = 1 × cos²(30°) = 0.75. So, the intensity is 0.75 units at that position.
Practical Applications, Common Pitfalls, and Links for Intensity In Young's Double Slit Experiment
YDSE intensity calculations are vital in JEE because they bridge theory and real measurement. Be careful with units and phase difference calculation. Only coherent sources give lasting fringes. In weighty numericals, always express positions using the fringe width formula for clarity.
- Intensity in Young's double slit experiment focused explanation
- Young's double slit experiment derivation for stepwise interference proofs
- Coherent and incoherent sources for real and ideal cases
- Huygens principle—wavefront explanation for interference
- Oscillations and waves for underlying wave behavior
- Optics mock test 1 for exam-style practice
- Optics revision notes for last-minute review
- Oscillations and waves practice paper for in-depth question sets
| Formula | Quantity | Meaning |
|---|---|---|
| I = I0 cos²(ϕ/2) | Intensity at phase ϕ | I0 = max intensity |
| Imax = 4I0 | Maximum | Central bright fringe |
| Imin = 0 | Minimum | Dark fringe |
| ϕ = (2π/λ)(d sinθ) | Phase difference | d = slit separation, λ = wavelength |
Consistent practice and strong conceptual clarity on intensity in Young's double slit experiment will help you tackle both direct formula questions and twist-based application problems in JEE. Bookmark this Vedantu summary for fast revision before the exam.
FAQs on Intensity Pattern in Young's Double Slit Experiment
1. What is the formula for intensity in Young's double slit experiment?
The intensity in Young's double slit experiment (YDSE) is given by: I = I₀ cos²(ϕ/2), where I is the intensity at a point on the screen, I₀ is the maximum intensity, and ϕ is the phase difference between the light from the two slits.
Key points:
- The formula shows how intensity varies with phase difference.
- Bright fringes (maxima) occur where cos²(ϕ/2) is maximum.
- Dark fringes (minima) occur where cos²(ϕ/2) is zero.
2. How is maximum intensity calculated in YDSE?
Maximum intensity (Imax) in Young's double slit experiment occurs when the phase difference ϕ = 0 or multiples of 2π (constructive interference).
Calculation:
- At maxima, Imax = 4I0, where I0 is the intensity from one slit.
- Occurs at points where path difference = mλ (m = 0, ±1, ±2...)
3. What is the intensity of Young's double slit experiment?
The intensity in Young's double slit experiment refers to the brightness observed at various points on the screen due to interference of light from two slits.
Summary:
- It varies periodically across the screen, forming alternating bright and dark fringes.
- Intensity depends on the phase or path difference between light waves from the two slits.
- Given by the distribution: I = I₀ cos²(ϕ/2), with maxima and minima at different positions.
4. Why does intensity vary across the screen in YDSE?
Intensity varies across the YDSE screen because of the constructive and destructive interference between waves from the two slits.
Main reasons:
- Points where the path difference is an integer multiple of the wavelength form bright fringes (high intensity).
- Half-integral multiples of wavelength lead to dark fringes (zero intensity).
- This alternating pattern creates the observed intensity variation across the screen.
5. What causes the bright and dark fringes in YDSE?
Bright and dark fringes in YDSE are caused by constructive and destructive interference of light waves from two coherent sources.
Details:
- Bright fringes: Occur where the path difference is nλ (n = 0, 1, 2,...); the amplitudes add up.
- Dark fringes: Occur where the path difference is (n + ½)λ; the amplitudes cancel each other.
6. How to find intensity at a point between two maxima in YDSE?
The intensity at any intermediate point between two maxima in YDSE is found using the formula: I = I₀ cos²(ϕ/2), where ϕ is the phase difference for that position.
Steps:
- Calculate the path difference at the desired point.
- Find the corresponding phase difference: ϕ = (2π/λ) × path difference.
- Substitute ϕ into the intensity formula to get the required value.
7. What is the intensity profile of a double slit?
The intensity profile of a double slit shows a periodic variation of brightness across the screen, featuring a series of bright and dark fringes.
Profile characteristics:
- Mathematically represented by I = I₀ cos²(ϕ/2).
- Maxima and minima correspond to constructive and destructive interference.
- The pattern is symmetric about the central maximum and repeats at equal intervals (fringe width).
8. Why does intensity decrease in double slits?
Intensity can decrease in double slits due to factors such as slit separation, finite slit width, or partial incoherence of light.
Causes:
- Diffraction effects from finite-width slits can spread out intensity, reducing maxima height.
- If slit widths are unequal, the resultant intensity at maximum is less than 4I₀.
- Loss of coherence in the sources results in reduced or washed-out fringe contrast.
9. Can intensity at a minimum ever be non-zero in YDSE?
The intensity at a minimum can be non-zero if the two slits have unequal intensities or are not perfectly coherent.
Key facts:
- Ideally, destructive interference (perfectly equal slits) gives zero intensity minima.
- If slit brightness differs, minima have some residual intensity (not zero).
- Imperfect coherence also causes minima to appear brighter than expected.
10. How does changing the wavelength affect the intensity pattern in YDSE?
Changing the wavelength in YDSE alters the fringe spacing (fringe width) but does not affect the intensity maxima value itself.
Effects:
- Longer wavelength → Wider fringe separation on screen.
- Shorter wavelength → Closer fringes.
- Maximum and minimum intensity values stay constant if slit widths and source intensity remain the same.
