Why cannot two independent, identical laser sources produce sustained interference fringes?

Independent light sources lack temporal coherence. Photons emitted by separate atomic transitions undergo random, uncorrelated phase jumps every $\sim 10^{-8}\text{ s}$. As a result, the phase difference $\Delta \phi$ varies rapidly with time, averaging out the cosine interference term $\cos(\Delta\phi) = 0$ and producing a uniform intensity $I_{\text{net}} = I_1 + I_2$ on screen.

How do I rapidly decide whether to add or subtract $(\mu-1)t$ in optical path equations?

Placing a transparent thin slab of thickness $t$ in front of a slit increases that ray's optical path by $(\mu-1)t$. The entire interference pattern shifts spatially toward the slit containing the slab. Always write path difference at height $y$ as $\Delta x = \left|\frac{y d}{D} - (\mu-1)t\right|$ if the slab is in the path of the upper slit.

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Question #1NEET 2024

In a Young's double-slit experiment, the separation between slits is $d = 0.5\text{ mm}$ and the screen distance is $D = 1\text{ m}$ . If light of wavelength $\lambda = 600\text{ nm}$ is used, calculate the fringe width $\beta$ .

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Step-by-Step Solution & Concept

Step 1

Convert all input parameters into standard SI units: $\lambda = 600 \times 10^{-9}\text{ m}$ , $D = 1\text{ m}$ , and $d = 0.5 \times 10^{-3}\text{ m}$ .

Step 2

Apply the fringe width formula $\beta = \frac{\lambda D}{d} = \frac{600 \times 10^{-9} \times 1}{0.5 \times 10^{-3}}$ .

Step 3

Simplify the expression: $\beta = 1200 \times 10^{-6}\text{ m} = 1.2 \times 10^{-3}\text{ m} = 1.2\text{ mm}$ .

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Question #2JEE Main 2023

In YDSE, monochromatic light of wavelength $\lambda_1 = 500\text{ nm}$ produces bright fringes. When replaced by another source of wavelength $\lambda_2$ , the $5^{\text{th}}$ bright fringe of $\lambda_1$ coincides with the $4^{\text{th}}$ bright fringe of $\lambda_2$ . Find $\lambda_2$ .

Solution Locked

Step-by-Step Solution & Concept

Step 1

Write the position formula for the $n^{\text{th}}$ bright fringe on the screen: $y_n = \frac{n \lambda D}{d}$ .

Step 2

Set the position condition $y_5(\lambda_1) = y_4(\lambda_2)$ , yielding $5 \frac{\lambda_1 D}{d} = 4 \frac{\lambda_2 D}{d} \implies 5\lambda_1 = 4\lambda_2$ .

Step 3

Substitute $\lambda_1 = 500\text{ nm}$ to solve for $\lambda_2$ : $\lambda_2 = \frac{5 \times 500}{4} = 625\text{ nm}$ .

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Mastering Wave Optics: Formulas, Traps & Exam Strategy | ExamiFly

Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Fringe Width in YDSE

Optical Path Length Shift

Angular Width of Central Maximum (Diffraction)

Brewster's Law

Interference Intensity Profile

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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