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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Light Waves

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Light Waves

In a Young's double slit experiment, two narrow vertical slits...

In a Young's double slit experiment, two narrow vertical slits positioned 0.800 mm0.800 \text{ mm} apart are illuminated by the same source of yellow light having a wavelength of 589 nm589 \text{ nm}. Determine the distance between adjacent bright bands in the interference pattern formed on a screen 2.00 m2.00 \text{ m} away.

Options

  1. A1.47 mm1.47 \text{ mm}
  2. B0.736 mm0.736 \text{ mm}
  3. C1.47 cm1.47 \text{ cm}
  4. D2.94 mm2.94 \text{ mm}

Correct answer

A. 1.47 mm1.47 \text{ mm}

Step-by-step solution

Given,

Wavelength of light, λ=589 nm=589×10−9 m\lambda = 589 \text{ nm} = 589 \times 10^{-9} \text{ m}

Distance between the slits, d=0.800 mm=0.800×10−3 md = 0.800 \text{ mm} = 0.800 \times 10^{-3} \text{ m}

Distance of the screen from the slits, D=2.00 mD = 2.00 \text{ m}

The distance between adjacent bright bands is the fringe width, given by the formula:

β=λDd\beta = \dfrac{\lambda D}{d}

Substituting the given values:

β=589×10−9×2.000.800×10−3\beta = \dfrac{589 \times 10^{-9} \times 2.00}{0.800 \times 10^{-3}}

β=1178×10−90.800×10−3\beta = \dfrac{1178 \times 10^{-9}}{0.800 \times 10^{-3}}

β=1472.5×10−6 m\beta = 1472.5 \times 10^{-6} \text{ m}

β=1.4725×10−3 m\beta = 1.4725 \times 10^{-3} \text{ m}

β≈1.47 mm\beta \approx 1.47 \text{ mm}
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