A monochromatic light of frequency 500 THz is incident on the slits of Young's double slit experiment. If the

Question

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:

Accepted Answer

1. **Identify Given Values:**
   - Frequency ($
u$) = $500 	ext{ THz} = 500 	imes 10^{12} 	ext{ Hz} = 5 	imes 10^{14} 	ext{ Hz}$.
   - Slit separation ($d$) = $0.2 	ext{ mm} = 0.2 	imes 10^{-3} 	ext{ m}$.
   - Screen distance ($D$) = $1 	ext{ m}$.
   - Number of fringes ($n$) = 10.
   - Speed of light ($c$) = $3 	imes 10^8 	ext{ m/s}$.
2. **Calculate Wavelength ($\lambda$):**
   $$ \lambda = \frac{c}{
u} = \frac{3 	imes 10^8}{5 	imes 10^{14}} = 0.6 	imes 10^{-6} 	ext{ m} = 6 	imes 10^{-7} 	ext{ m} $$
3. **Calculate Fringe Width ($\beta$):**
   The formula for fringe width is $\beta = \frac{\lambda D}{d}$.
   $$ \beta = \frac{(6 	imes 10^{-7} 	ext{ m})(1 	ext{ m})}{0.2 	imes 10^{-3} 	ext{ m}} $$
   $$ \beta = \frac{6 	imes 10^{-7}}{2 	imes 10^{-4}} = 3 	imes 10^{-3} 	ext{ m} = 3 	ext{ mm} $$
4. **Calculate Width of 10 Fringes:**
   $$ 	ext{Width} = n 	imes \beta = 10 	imes 3 	ext{ mm} = 30 	ext{ mm} $$

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