Question

Which one of the following figures correctly depicts the intensity distribution for Fraunhofer diffraction due to a single slit? Here, x denotes the distance from the centre of the central fringe and I denotes the intensity.

• A.

  

• B.

  

• C.

  

• D.

  

Answer 1

The Correct Option is C

To solve this question, we need to understand the concept of Fraunhofer diffraction due to a single slit. In such a diffraction pattern, the intensity distribution is characterized by a central maximum that is much brighter and wider than the subsequent minima and side fringes.

The intensity \(I(\theta)\) as a function of the angle \(\theta\) (or equivalently, the position \(x\) on the screen) is given by:

\(I(\theta) = I_0 \left( \frac{\sin(\beta)}{\beta} \right)^2\)

where \(\beta = \frac{\pi a}{\lambda} \sin \theta\), \(a\) is the slit width, and \(\lambda\) is the wavelength of the light.

 

This function results in a central maximum (principal maximum) with intensity \(I_0\) and diminishing secondary maxima (side lobes) on either side separated by dark fringes (minima).

Let us justify why the given correct image depicts this pattern accurately:

1. Central maximum: The central peak is the highest and widest, consistent with the property of Fraunhofer diffraction.
2. Side lobes: The side fringes decrease in intensity as we move away from the center, in accordance with the expression \(\left( \frac{\sin(\beta)}{\beta} \right)^2\).
3. Pattern symmetry: The pattern is symmetric about the central maximum, as expected from a single slit.

Therefore, the correct figure illustrating the intensity distribution for Fraunhofer diffraction due to a single slit is selected based on these characteristics.

By analyzing these features, we can confirm that the selected figure accurately represents the intensity distribution of light for single-slit Fraunhofer diffraction.