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A monochromatic light of frequency 500 THz is incident on the slits of Young's double slit experiment. If the distance between the slits is 0.2 mm and the screen is placed at a distance 1 m from the slits, the width of 10 fringes will be:

A

1.5 mm

B

15 mm

C

30 mm

D

3 mm

Step-by-Step Solution

  1. Identify Given Values:
  • Frequency (ν\nu) = 500 THz=500×1012 Hz=5×1014 Hz500 \text{ THz} = 500 \times 10^{12} \text{ Hz} = 5 \times 10^{14} \text{ Hz}.
  • Slit separation (dd) = 0.2 mm=0.2×10−3 m0.2 \text{ mm} = 0.2 \times 10^{-3} \text{ m}.
  • Screen distance (DD) = 1 m1 \text{ m}.
  • Number of fringes (nn) = 10.
  • Speed of light (cc) = 3×108 m/s3 \times 10^8 \text{ m/s}.
  1. Calculate Wavelength (λ\lambda): λ=cν=3×1085×1014=0.6×10−6 m=6×10−7 m\lambda = \frac{c}{\nu} = \frac{3 \times 10^8}{5 \times 10^{14}} = 0.6 \times 10^{-6} \text{ m} = 6 \times 10^{-7} \text{ m}
  2. Calculate Fringe Width (β\beta): The formula for fringe width is β=λDd\beta = \frac{\lambda D}{d}. β=(6×10−7 m)(1 m)0.2×10−3 m\beta = \frac{(6 \times 10^{-7} \text{ m})(1 \text{ m})}{0.2 \times 10^{-3} \text{ m}} β=6×10−72×10−4=3×10−3 m=3 mm\beta = \frac{6 \times 10^{-7}}{2 \times 10^{-4}} = 3 \times 10^{-3} \text{ m} = 3 \text{ mm}
  3. Calculate Width of 10 Fringes: Width=n×β=10×3 mm=30 mm\text{Width} = n \times \beta = 10 \times 3 \text{ mm} = 30 \text{ mm}
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