Diffraction and Interference: Single and Double Slits

When a slit is made wider, the diffraction pattern observed becomes narrower, indicating that the waves pass through the slit with less bending and hence less dispersion of light. This is because wider slits allow more wavefronts to pass through simultaneously, reducing the angle at which the lightwaves spread out beyond the slit location. As a result, the fringe separation between the bright and dark bands on the screen decreases, leading to a more concentrated pattern.
Conversely, as the slit width decreases, the light waves spread out more, producing a wider fringe pattern on the screen. A narrower slit restricts the amount of light entering, leading to greater diffraction effects where the light beams diverge at larger angles.

Given:
- Wavelength (\lambda) = 640 \text{ nm} = 640 \times 10^{-9} \text{ m}
- Slit width (d) = 0.08 \text{ mm} = 0.08 \times 10^{-3} \text{ m}
- Distance to the screen (D) = 1.2 \text{ m}

Formula: \theta = \frac{\lambda}{d}
Calculation: \theta = \frac{640 \times 10^{-9}}{0.08 \times 10^{-3}} = 0.008 \text{ radians}

Formula: \tan(\theta) = \frac{s}{D}, which can be approximated to \theta = \frac{s}{D} when \theta is small.
Where:
- s is the fringe distance (half the width of the central maximum).
- D is the distance from the slit to the screen.
Solving for s: s = \theta \times D = 0.008 \times 1.2 = 0.0096 \text{ m}
The full width of the central maximum is thus calculated as 2s = 2 \times 0.0096 = 0.0192 \text{ m} = 19.2 \text{ mm}, indicating the size of the central bright fringe observed in diffraction patterns.

Given:
- Slit separation (d) = 0.5 \text{ mm} = 0.5 \times 10^{-3} \text{ m}
- Distance to the screen (D) = 2.4 \text{ m}
- Fringe separation (s) = 2.84 \text{ mm} = 2.84 \times 10^{-3} \text{ m}

Formula: d \sin(\theta) = n \lambda, where n = 1 (first bright fringe). To simplify calculations for small angles, we can approximate this to be d \frac{s}{D} = \lambda.
Calculation: \lambda = \frac{d \times s}{D} = \frac{0.5 \times 10^{-3} \times 2.84 \times 10^{-3}}{2.4} = 5.9167 \times 10^{-7} \text{ m} = 591.67 \text{ nm}, which indicates the wavelength of light used in the experiment.

Given: New wavelength (\lambda) = 468 \text{ nm} = 468 \times 10^{-9} \text{ m}
Formula: s = \frac{n \lambda D}{d}, where n is the order of the fringe.
Calculation: s = \frac{1 \times 468 \times 10^{-9} \times 2.4}{0.5 \times 10^{-3}} = 0.0022464 \text{ m} = 2.2464 \text{ mm}. This result shows how a change in wavelength directly affects the separation between the bright fringes in the diffraction pattern.

Single Slit Diffraction:
- Angle for the first minimum: \theta = \frac{\lambda}{d}, indicating how slit width influences diffraction patterns.
Double Slit Interference:
- Fringe separation: s = \frac{n \lambda D}{d}, highlights the relationship between fringe distance, wavelength, and slit separation, crucial in understanding interference patterns.

Start with questions 8 and 9, then move on to 11 and 12, to deepen understanding of diffraction and interference concepts.