Question

In the diffraction from a single slit of width $2.5\lambda$, the total number of minima and secondary maximum (maxima) on either side of the central maximum are:
A) 4 minimas, 2 secondary maximas
B) 2 minimas, 2 secondary maximas
C) 4 minimas, 4 secondary maximas
D) 4 minimas, 3 secondary maximas

Answer 1

In a single slit diffraction experiment, the light from one portion of the slit can interfere with light from another portion and the resultant intensity on the screen will depend upon whether interference is constructive or destructive. It is found that the diffraction pattern on the screen consists of a central bright band called a central maximum having alternating dark and less bright bands of decreasing intensity on both sides of the central maximum.

Complete step by step answer:
It is given that the width of the slit, $a = 2.5\lambda$
We know that, for minima
$a\sin \theta = n\lambda$
Taking $\sin \theta$ as LHS and remaining terms as in RHS we get,
$\sin \theta = \dfrac{{n\lambda }}{a}$
Substitute the value of$a$ so we can write it as,
$\sin \theta = \dfrac{{n\lambda }}{{2.5\lambda }}$
Cancel the same term we get,
$\rightarrow \sin \theta = \dfrac{n}{{2.5}}$
Since the maximum value of$\sin \theta = 1$
So, only$n{\text{ }} = {\text{ }}1,{\text{ }}2$,
Thus, only $4$ minima can be obtained on both sides of the screen.
For secondary maxima,
$a\sin \theta = \left( {2n + 1} \right)\dfrac{\lambda }{2}$
Taking $\sin \theta$as LHS and remaining terms as in RHS we get,
$\sin \theta = \dfrac{{\left( {2n + 1} \right)\lambda }}{{2a}}$
Substitute the value of$a$ we get
$\sin \theta = \dfrac{{\left( {2n + 1} \right)\lambda }}{{2 \times 2.5\lambda }}$
Cancelling the same term we get,
$\rightarrow \sin \theta = \dfrac{{\left( {2n + 1} \right)}}{5}$
Since the maximum value of $\sin \theta = 1$.
So, only n = 1, thus, only 2 secondary maxima can be obtained on both sides of the screen.

therefore Correct option is A

Additional information:
-The phenomenon of bending of light around the edges of an obstacle is called diffraction of light.
-Due to the bending of light around the corners of an obstacle, the light deviates from its straight-line path and enters into the geometrical shadow of the obstacle.
-Examples of diffraction of light:

1. At the time of a solar eclipse. Shadow bands are seen on the earth due to diffraction of sunlight.
2. The images of stars in a telescope do not appear as a sharp point but appear as diffused spots due to diffraction.

Note:
-Diffraction is due to the superposition of secondary wavelets coming from different points of the same wavefronts.
-In the diffraction pattern, the intensity of successive bright fringes goes on decreasing.
-Never diffraction fringes of the same width
-In diffraction, bands are a few in numbers.