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If the distance between the slits in YDSE is halved and the screen distance is doubled, the fringe width becomes:

4 times. Fringe width β = λD/d. Replacing D with 2D and d with d/2 gives β' = λ(2D)/(d/2) = 4(λD/d) = 4β.

Physics · Class 12 · JEE / NEET / AP Physics

Wave Optics & Young's Double Slit Experiment

Wave Opticsvalidates the wave nature of light. InYoung's Double Slit Experiment (YDSE), two coherent secondary wavefronts overlap to generate alternating bright and dark interference fringes on a distant observation screen.

11. Visual Overview & Intuitive Discovery: Wave Optics & Young's Double Slit Experiment

In 1801, Thomas Young performed a historic experiment with sunlight, a pinhole, and two narrow slits in a darkened room. What he saw shocked the scientific world: where two rays of light overlapped on a screen, they created dark bands of absolute blackness! Adding more light created darkness. This proved beyond doubt that light is a wave, capable of destructive self-cancellation.

Young's Double Slit Interference Setup: Fringe Width (β), Constructive Interference, Destructive Interference, Medium Immersion Effect — Young's Double Slit Interference Setup
Wavefront interference geometry with slit spacing d, screen distance D, and fringe width β = λD/d.
1Fringe Width (β) — β = λ · D / d (spacing between two consecutive bright or dark bands)
2Constructive Interference — Path difference Δx = n λ, producing maximum intensity I_max = 4 I₀
3Destructive Interference — Path difference Δx = (2n - 1) λ / 2, producing zero intensity dark fringes
4Medium Immersion Effect — In a liquid of index μ, wavelength shrinks to λ/μ, reducing fringe width to β/μ

Interactive Study Prompts (Click to Explore)

Clickable

AI Deep Dive 🪄

22. Physical Mechanism Flowchart: Wave Optics & Young's Double Slit Experiment

Wave Optics & Young's Double Slit Experiment Step-by-Step Flowchart

1
Initial State
Initial parameters and boundary conditions are established.
2
Fundamental Force/Interaction
Governing physical laws act continuously across the system.
3
Energy & Vector Conservation
Conservation principles restrict permitted trajectory states.
4
Dynamic Equilibrium
Final observable state satisfies equilibrium equations.

Follow the numbered pipeline above. In physics examinations, sketching this mental flowchart ensures you never miss intermediate steps in complex multi-concept numericals.

33. Rigorous Mathematical Proofs & Governing Equations

Derivation — Derivation of Governing Formula for Wave Optics & Young's Double Slit Experiment

1
$F_{\text{net}} = m a$
Starting from Newton's foundational law
2
$\int F \, dx = \Delta K$
Integrating over displacement

Resultant Intensity Formula

I = 4 I_0 \cos^2\left(\frac{\phi}{2}\right),\quad \phi = \frac{2\pi}{\lambda}\Delta x

$I_0$ = Intensity from an individual individual slit
$\phi$ = Phase difference between overlapping light waves

Core Physical Variables

Variable	Symbol	SI Unit
Energy	E	Joules (J)
Power	P	Watts (W)

Energy
Symbol
E
SI Unit
Joules (J)
Power
Symbol
P
SI Unit
Watts (W)

44. Nanotechnology: Antireflective Coatings on Camera Lenses

High-end camera lenses, microscope objectives, and eyeglass lenses have a faint purplish or green sheen. This is thin-film destructive wave interference! Optical manufacturers apply a microscopic magnesium fluoride ( $MgF_2$ ) coating exactly one-quarter of a wavelength thick ( $\lambda / 4$ ). Light reflecting from the top surface of the coating interferes destructively with light reflecting from the glass interface beneath it, eliminating surface reflections from 4% down to under 0.1%, channeling 99.9% of light into the sensor.

Exam tip: Conservation of Energy in Interference: Destructive interference does NOT destroy or extinguish light energy! The energy that disappears from dark fringes is redistributed into the bright fringes, where peak intensity (

4I_0

) is four times that of an individual slit, maintaining perfect total energy conservation across the screen.

55. Solved Standard & Advanced Exam Problems

Exam question · JEE Standard

Sample Question

Exam question · JEE Advanced / AP Physics

In a Young's double slit experiment using light of wavelength λ, the intensity at a point where the path difference is λ/6 is what fraction of the maximum intensity I_max?

A. $3/4 I_max$
B. $1/2 I_max$
C. $1/4 I_max$
D. $1/3 I_max$

Remember

In YDSE, all bright and dark fringes have the exact same width: β = λD/d.
Energy is not destroyed in destructive interference; it is redistributed into bright interference zones.
If the double slit setup is submerged in water, fringe width decreases because wavelength shortens.

Check yourself

Question 1
If the distance between the slits in YDSE is halved and the screen distance is doubled, the fringe width becomes: