Question

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 be\(x \times 10^{–5}\)m. The value of x is :

Answer 1

Correct Answer: 275

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

The distance \( y \) to the first order minima in a single-slit diffraction pattern is given by:

\[ y = \frac{\lambda D}{d}. \]

Substitution
Substituting the given values:

\[ y = \frac{550 \times 10^{-9} \times 1}{0.2 \times 10^{-3}}. \]

Calculation
Simplifying:

\[ y = \frac{550 \times 10^{-9} \times 10^2}{0.2 \times 10^{-3}} = \frac{550 \times 10^{-7}}{0.2 \times 10^{-3}}. \]

Further simplification:

\[ y = \frac{550 \times 10^{-5}}{0.2} = 275 \times 10^{-5} \, \text{m}. \]

Therefore, the value of \( x \) is 275.