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JEE Main 201212 MayOnline
Exam
JEE Main
Year
2012
Date
12 May
Mode
Online
Subject
Physics
Chapter
Wave Optics

JEE Main 2012 · 12 May · Online · Physics · Wave Optics

The maximum number of possible interference maxima for slit...

The maximum number of possible interference maxima for slit separation equal to 1.8λ1.8 \lambda, where λ\lambda is the wavelength of light used, in a Young's double slit experiment is

Options

  1. Azero
  2. B3
  3. Cinfinite
  4. D5

Correct answer

B. 3

Step-by-step solution

As sin⁡θ=nλd\sin \theta=\frac{n \lambda}{d} and sin⁡θ\sin \theta cannot be ≯1\ngtr 1 \therefore 1=nλ1.8λ\frac{n \lambda}{1.8 \lambda}or n=1.8n=1.8 Hence maximum number of possible interference maximas, 0,±10, \pm 1 i.e. 3
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