prove that in single slit experiment the width of secondary maxima is half the width of central maxima?
from ch wave optics
Answers
Solution :
Let the width of the slit be 'a' when this is doubled then the half angular width of central maximum which is λa, reduces to half .
Hence the intensity of central maximum will become 4 times , because the area of central diffraction band would be 1/4 th .
Answer : intensity of central diffraction band will become 4 times.
In single slit Fraunhofer diffraction, why is the width of the central maximum twice the width of the other fringes?
To understand this properly requires the use of mathematics, but I will give you a simple answer, without much maths.
If you look at diagrams of Fraunhofer single-slit diffraction patterns, you will see that the central maximum is symmetrical about the centre line. All the other maxima and minima are found on just one side of the centre line. Of course, the pattern as a whole is also symmetrical about the centre line.
Both the mathematics and experimental results show that the positions of the minima are at distances from the centre line that depend on the details of the particular experiment.
Those minima distances are found fairly accurately by multiplying a constant (which depends on the experiment) by a whole number (that number is 1 for the first two minima and 2 for the second two minima, and so on). There is no minimum on the centre line, of course (corresponding to the number 0).
Let Min1 be the distance from the centre line to the first two minima, one on each side of the centre line. Let the distance of the second two minima from the centre line be Min2, and so on. Min2 is about 2 x Min1. The width of the maxima is roughly the distance between adjacent minima. So, all the secondary maxima are about Min1 in width.
However, the distance between the two first two minima, on each side of the centre line is 2 x Min1. The central maximum lies between the first two minima (located on opposite sides of the centre line). All the secondary bright fringes lie between adjacent minima on the same side of the centre line. So now you can see why the width of the central maximum is about twice the width of the other bright fringes.