Step-by-Step Guide to Young’s Double Slit Experiment Derivation

The Young’s Double Slit Experiment Derivation is foundational for understanding interference of light, a central topic in JEE Main Physics. Thomas Young’s experiment showcased that light shows wave behaviour by producing distinctive bright and dark fringes. Aspirants must master how the interference pattern arises, the stepwise logic of the derivation, and the equations governing fringe positions and widths.

Physical Setup and Principle of Young’s Double Slit Experiment Derivation

In the double slit experiment, a monochromatic light source is allowed to pass through two narrow, parallel slits separated by distance d. The light emerging from the slits travels towards a distant screen kept at distance L (with L ≫ d). Since both slits originate from the same source, they maintain a constant phase relationship, qualifying as coherent sources.

The waves from each slit travel slightly different distances to each point on the screen, causing constructive or destructive interference depending on the path difference. This leads to the formation of alternate bright and dark bands, called fringes, along the screen.

Complete Derivation of Young’s Double Slit Condition and Fringe Formulae

Let us logically derive key results for the interference pattern. Follow each step precisely for full marks in JEE Main:

Step 1: Path difference expression for a point P at vertical distance y from the central axis:
Let the two slits be S₁ and S₂, separated by d, and screen placed at distance L. The path difference for point P is given by:
Δx = S₂P − S₁P
For small angles (since L ≫ d),
sin θ ≈ tan θ = y / L
So, Δx = d sin θ = d y / L

Step 2: Conditions for constructive and destructive interference:
- Constructive (bright fringe): Path difference = mλ, where m = 0, ±1, ±2…
Setting Δx = mλ gives:
d y / L = mλ
So, y_bright = (mλL) / d
- Destructive (dark fringe): Path difference = (m + ½)λ
d y / L = (m + ½)λ
So, y_dark = [(m + ½)λL] / d

Step 3: Fringe width calculation (distance between two consecutive bright or dark fringes):
Fringe width, β = y_m+1 − y_m
= [(m + 1)λL / d] − [mλL / d]
= (λL) / d

Final Expression: The fringe width in Young’s Double Slit Experiment is β = λL / d.

Physical Meaning of Constructive and Destructive Interference

At some points on the screen, the path difference between waves from each slit equals an integer multiple of the wavelength, resulting in constructive interference (bright bands). Where the path difference is a half-integer multiple, destructive interference occurs, producing dark bands.

• Bright fringes appear where waves reinforce due to zero or integral-multiple path difference.
• Dark fringes result from anti-phase superposition, cancelling the light intensity.
• The central fringe at y = 0 is always bright, corresponding to m = 0.

Key Parameters Influencing Young’s Double Slit Experiment Derivation

According to the derivation, the clarity and separation of fringes depend on several factors. Understanding their effects helps solve JEE questions efficiently.

• Increasing slit separation d decreases fringe width (β).
• Larger L increases spread of fringes on the screen.
• Shorter wavelength light yields finer, closely spaced fringes.
• Coherence of sources is essential; lack of coherence destroys the interference pattern.

Accuracy and Assumptions in Practical JEE Problems

For JEE Main, problems on Young’s double slit experiment may involve changes in medium, slit separation, or source wavelength. Always remember the small-angle approximation (sin θ ≈ tan θ ≈ y / L) is valid for typical setups where the screen is significantly farther than slit spacing.

Additionally, calculations assume narrow slits, equal intensity from both, and negligible diffraction effects. Complex cases, like introducing a transparent material or varying λ, require careful substitution into derived formulae for correct results.

Vedantu’s JEE platform provides crisp, stepwise explanations matching NCERT and exam-oriented expectations for Young's Double Slit Experiment Derivation, ensuring you can solve all related pattern and calculation problems easily.