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JEE Main 20207 JanuaryEvening Shift
Exam
JEE Main
Year
2020
Date
7 January
Shift
Evening Shift
Subject
Physics
Chapter
Wave Optics

JEE Main 2020 · 7 January · Evening Shift · Physics · Wave Optics

In a Young’s double slit experiment, the separation between...

In a Young’s double slit experiment, the separation between the slits is 0.15mm0.15mm . In the experiment, a source of light of wavelength 589nm589nm is used and the interference pattern is observed on a screen kept 1.5m1.5m away. The separation between the successive bright fringes on the screen is:

Options

  1. A6.9mm6.9mm
  2. B3.9mm3.9mm
  3. C5.9mm5.9mm
  4. D4.9mm4.9mm

Correct answer

C. 5.9mm5.9mm

Step-by-step solution

β=\lambdaDd=589×10−9×1.50.15×10−3=5.9mm\beta = \dfrac{\lambdaD}{d} = \dfrac{589 \times {10}^{- 9} \times 1.5}{0.15 \times {10}^{- 3}} = 5.9mm

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