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Physics Question on Wave optics

In a single slit experiment, a parallel beam of green light of wavelength 550 nm passes through a slit of width 0.20 mm. The transmitted light is collected on a screen 100 cm away. The distance of first order minima from the central maximum will bex×10–5x \times 10^{–5}m. The value of x is :

Answer

Given data:
- Wavelength of light, λ=550 nm=550×10−9 m\lambda = 550 \, \text{nm} = 550 \times 10^{-9} \, \text{m}
- Distance to the screen, D=100 cm=1 mD = 100 \, \text{cm} = 1 \, \text{m}
- Width of the slit, d=0.2 mm=0.2×10−3 md = 0.2 \, \text{mm} = 0.2 \times 10^{-3} \, \text{m}

The distance yy to the first order minima in a single-slit diffraction pattern is given by:

y=λDd.y = \frac{\lambda D}{d}.

Substitution
Substituting the given values:

y=550×10−9×10.2×10−3.y = \frac{550 \times 10^{-9} \times 1}{0.2 \times 10^{-3}}.

Calculation
Simplifying:

y=550×10−9×1020.2×10−3=550×10−70.2×10−3.y = \frac{550 \times 10^{-9} \times 10^2}{0.2 \times 10^{-3}} = \frac{550 \times 10^{-7}}{0.2 \times 10^{-3}}.

Further simplification:

y=550×10−50.2=275×10−5 m.y = \frac{550 \times 10^{-5}}{0.2} = 275 \times 10^{-5} \, \text{m}.

Therefore, the value of xx is 275.