Inactive Tutor answered • 04/04/24
To find the distance between the central maximum and the first place where the intensity is zero due to the single slit effect in a diffraction pattern, we can use the formula for the angular position of the minima in single-slit diffraction:
\[ \theta = \sin^{-1} \left( \frac{m \lambda}{a} \right) \]
where:
- \( \theta \) is the angular position of the minimum,
- \( m \) is the order of the minimum (for the first minimum, \( m = 1 \)),
- \( \lambda \) is the wavelength of light,
- \( a \) is the width of the slit.
The distance \( x \) between the central maximum and the first minimum can be calculated using the relation between the angle \( \theta \) and the distance \( x \) from the slit to the screen:
\[ x = a \tan \theta \]
Given that you're asking about the first minimum, \( m = 1 \), we can proceed with the calculations.
Let's assume:
- Wavelength of light \( \lambda = 600 \) nm (this is a common value for visible light),
- Width of the slit \( a = 0.2 \) mm = \( 0.2 \times 10^{-3} \) m.
First, calculate the angle \( \theta \) for the first minimum using the formula:
\[ \theta = \sin^{-1} \left( \frac{m \lambda}{a} \right) \]
\[ \theta = \sin^{-1} \left( \frac{1 \times 600 \times 10^{-9}}{0.2 \times 10^{-3}} \right) \]
\[ \theta = \sin^{-1} (0.003) \]
Now, calculate the distance \( x \) using \( x = a \tan \theta \):
\[ x = 0.2 \times 10^{-3} \times \tan(\sin^{-1}(0.003)) \]
Calculating this value gives:
\[ x \approx 0.002 \, \text{m} \]
So, the distance between the central maximum in the diffraction pattern and the first place where the intensity is zero due to the single-slit effect is approximately 0.002 meters.
