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.

Answer 2

Step 1: Given data.
Wavelength of light, λ = 550 nm = 550 × 10⁻⁹ m
Width of slit, a = 0.20 mm = 2 × 10⁻⁴ m
Distance between slit and screen, D = 100 cm = 1 m

Step 2: Formula for the position of the first order minimum.
For a single slit diffraction pattern, the condition for minima is:
\[ a \sin \theta = m \lambda \] where m = ±1, ±2, ±3 ... (order of minima).

For small angles, \( \sin \theta ≈ \tan \theta = \frac{y}{D} \), where y is the distance of the fringe from the central maximum.
Hence, \[ a \frac{y}{D} = m \lambda \] For first order minima (m = 1):
\[ y = \frac{D \lambda}{a} \]

Step 3: Substitute the values.
\[ y = \frac{1 \times 550 \times 10^{-9}}{2 \times 10^{-4}} = 2.75 \times 10^{-3} \, \text{m} \]
\[ y = 2.75 \times 10^{-3} \, \text{m} = 275 \times 10^{-5} \, \text{m} \]

Step 4: Final Answer.
The value of x is:
\[ \boxed{275} \]

Final Answer: 275