Question

Differentiate between interference and diffraction of light. Explain qualitatively the diffraction phenomenon of light by a single slit. Light of 6000 Ã… wavelength is incident normally on a single slit of width \( 3 \times 10^{-4} \, \text{cm} \). Find out the angular width of the central maxima. 
 

Hint:

Diffraction and interference demonstrate the wave nature of light. Diffraction is more evident with small apertures or obstacles.

Answer 1

Difference Between Interference and Diffraction:
Pattern Origin and Superposition of waves from coherent sources and Light bends.
Fringes: All equally bright; spaced.
Central fringe brighter and wider; others diminish.
Cause: Interaction of more light beams
Single light wave- edges of aperture
Single-Slit Diffraction: - When light passes through a single slit, it bends and forms a pattern of alternating bright and dark fringes on a screen.
- The intensity distribution is maximum at the center (central maxima) and decreases for higher-order fringes. - At specific angles, destructive interference occurs due to the path difference, forming dark fringes. \[ a \sin \theta = n\lambda, \, \text{where } a \text{ is the slit width, } \lambda \text{ is the wavelength, and } n = \pm 1, \pm 2, \dots \] 3. Angular Width of Central Maximum:
- Central maximum extends between \( -\theta \) and \( +\theta \). - The angular width \( \Delta \theta \) is: \[ \Delta \theta = 2\sin^{-1} \left(\frac{\lambda}{a}\right). \] Substituting \( \lambda = 6000 \, \text{Ã…} = 6 \times 10^{-5} \, \text{cm} \) and \( a = 3 \times 10^{-4} \, \text{cm} \): \[ \sin \theta = \frac{\lambda}{a} = \frac{6 \times 10^{-5}}{3 \times 10^{-4}} = 0.2. \] \[ \Delta \theta = 2\sin^{-1}(0.2) \approx 2 \times 11.54^\circ = 23.08^\circ. \] Final Answer:
The angular width of the central maxima is approximately \( 23.08^\circ \).