Young's Double Slit Experiment Derivation: Formula, Steps & Diagram

Young's Double Slit Experiment Derivation is a cornerstone of wave optics for JEE Main aspirants. It shows how light produces an interference pattern, and enables students to derive and apply the important fringe width formula. Mastery of this derivation is crucial for scoring well in topics related to interference and the wave nature of light.

Introduction to Young's Double Slit Experiment Derivation in JEE

Young's double slit experiment derivation demonstrates that light exhibits interference, supporting its wave nature. The experiment uses two narrow, closely spaced slits illuminated by a single source. The resulting pattern of bright and dark bands on a screen allows us to derive the equation for fringe width and understand conditions for maxima and minima.

This derivation features regularly in the JEE Main syllabus and frequently appears in practice tests and board exams. Understanding each step and the underlying physics is essential for tackling related problems and their variations.

Experimental Setup and Path Difference in Young's Double Slit Experiment Derivation

The experimental setup involves:

• A monochromatic light source placed behind a single slit
• Light incident on two parallel, close slits (S₁ and S₂) separated by distance d
• The slits act as coherent sources
• An observation screen at distance D from the slits, with D ≫ d
• A point P on the screen at vertical distance y from the central axis

The path difference between waves from S₁ and S₂ at point P is given by (using small angle approximation):

Path difference (Δ) ≈ (d · y) / D

A labelled diagram is vital for visualizing point P, slit separation d, and distances. Use references like coherence and coherent sources for further clarity.

Principle of Interference and Theory in Young's Double Slit Experiment Derivation

Young's double slit experiment derivation uses the principle of superposition of waves. When two coherent light waves overlap, their amplitudes add or subtract, forming bright and dark fringes due to constructive and destructive interference respectively.

• Bright fringes (maxima) occur when the path difference is an integral multiple of wavelength (mλ, where m = 0, 1, 2...)
• Dark fringes (minima) form for odd multiples of half-wavelength ((2m+1)λ/2)
• The screen shows an alternating interference pattern of light and dark bands

These ideas are central to the superposition of SHM and form the foundation for more advanced concepts like thin film interference.

Stepwise Young's Double Slit Experiment Derivation of Fringe Width Formula

The main goal in Young's double slit experiment derivation is the equation for fringe width (β). Follow these steps:

1. Let d = slit separation, D = screen distance, λ = wavelength.
2. Path difference at point P: Δ = (d · y) / D
3. Condition for constructive interference (bright fringe): Δ = mλ ⇒ y = (mλD)/d
4. Condition for destructive interference (dark fringe): Δ = (2m+1)λ/2 ⇒ y = ((2m+1)λD)/(2d)
5. The distance between two consecutive bright fringes (fringe width) is the difference between positions for m and m+1:

Fringe width, β = y_m+1 - y_m = (λD)/d

Remember, this β = λD/d formula is valid when D ≫ d and for monochromatic coherent sources. JEE exams often test derivation steps or calculations with this equation.

For a quick recap of interference patterns, refer to intensity in Young's double slit experiment and optics collections.

Bright and Dark Fringes: Locations and Pitfalls in Young's Double Slit Experiment Derivation

In Young's double slit experiment derivation, bright fringes appear at positions:

• y = (mλD)/d, where m = 0, ±1, ±2, ... (central bright at m = 0)

For dark fringes:

• y = ((2m+1)λD)/(2d), where m = 0, ±1, ...

Remember these key points:

• The central maximum is always a bright fringe
• Fringe widths for all bright (or all dark) fringes are equal
• For polychromatic (non-monochromatic) light, colored patterns overlap and fringe clarity reduces
• Errors in slit width or alignment cause fringe blurring

Review these ideas alongside polarisation of light and related wave motion concepts.

Applications, Numericals, and Key Revision: Young's Double Slit Experiment Derivation

To apply Young's double slit experiment derivation, try typical JEE numericals. Practice using actual fringe width values and changing D, d, or λ. Here's a classic example:

• A double slit arrangement uses λ = 600 nm, d = 0.2 mm, and D = 1 m. Find fringe width β.

1. Convert everything to SI: λ = 600 × 10^-9 m, d = 2 × 10^-4 m, D = 1 m.
2. Apply the formula: β = λD/d.
3. Calculate: β = (600 × 10^-9 × 1) / (2 × 10^-4) = 3.0 × 10^-3 m = 3 mm.

Practice more with oscillations and waves practice papers or mock tests for multiple question styles.

• Always check units and convert to metres before substitution
• Ensure distances in the setup meet the D ≫ d condition
• If any source is incoherent or emits multiple wavelengths, the derivation no longer holds cleanly

Vedantu’s JEE preparation resources feature advice for mastering fringe width derivation, optics, and error-free calculation.

For last-minute revision, keep these points in mind:

• Fringe width formula: β = λD/d
• Bright fringe locations: y = (mλD)/d
• Use only monochromatic, coherent sources
• If D or d changes, β changes accordingly; adjust in sums
• Review setups in principle of superposition of waves, coherence and Huygens principle

Understanding and practicing the Young's Double Slit Experiment Derivation builds deep conceptual clarity for JEE Main wave optics. For more concise notes and solved examples, download resources directly from the Vedantu study portal.