Question
Derive the conditions for bright and dark fringes produced due to diffraction by a single slit.
✓
Answer
When a parallel beam of monochromatic light of wavelength λ illuminates a single slit of finite width a, we observe on a screen some distance from the slit, a broad pattern of alternate dark and bright fringes. The pattern consists of a central bright fringe, with successive dark and bright fringes of diminishing intensity on both sides. This is called ‘ the diffraction pattern of a single slit.Consider a single slit illuminated with a parallel beam of monochromatic light perpendicular to the plane of the slit. The diffraction pattern is obtained on a screen at a distance D (» a) from the slit and at the focal plane of the convex lens, Fig. $7.33.$
We can imagine the single slit as being made up of a large number of Huygens' sources evenly distributed over the width of the slit. Then the maxima and minima of the pattern arise from the interference of the various Huygens' wavelets.
Now, imagine the single slit as made up of two adjacent slits, each of width $a / 2$. Since, the incident plane wavefronts are parallel to the plane of the slit, all the Huygens sources at the slit will be in phase. They will therefore also in phase at the point $P _0$ on the screen, where $P _0$ is equidistant from all the Huygens sources. At $P _0$, then, we get the central maximum.
For the first minimum of intensity on the screen, the path difference between the waves from the Huygens sources A and $O$ (or $O$ and $B$ ) is $\lambda / 2$, which is the condition for destructive interference. Suppose, the nodal line $O P$ for the first minimum subtends an angle $\theta$ at the slit; $\theta$ is very small. With $P$ as the centre and $P A$ as radius, strike an arc intersecting $P B$ at $C$. Since, $D \gg a$, the arc $A C$ can be considered a straight line at right angles to $P B$. Then, $\triangle A B C$ is a right-angled triangle similar to $\triangle OP P _0 P$.
This means that, $\angle BAC = \theta$
$\therefore BC = a \sin\theta$
$\therefore $ Difference in path length,
$BC = PB – PA = (PB – PO) + (PO – PA)$
$=\frac{\lambda}{2}+\frac{\lambda}{2}$
$=\lambda$ $ \therefore a \sin \theta=\lambda$
$\therefore \sin \theta \simeq \theta=\frac{\lambda}{a} $
( $\because \theta$ is very small and in radian)
The other nodal lines of intensity minima can be understood in a similar way. In general, then, for the with minimum ( $m= \pm 1, \pm 2, \pm 3, \ldots)$.
$\theta_m=\frac{m \lambda}{a}$ (mth minimum) $\ldots$ (2)
as $\theta_m$ is very small and in radian.
Between the successive minima, the intensity rises to secondary maxima when the path difference is an odd-integral multiple of $\frac{\lambda}{2}$ :
$a \sin \theta_m=(2 m+1) \frac{\lambda}{2}=\left(m+\frac{1}{2}\right) \lambda$
i.e., at angles given by,
$\theta_{ m } \simeq \sin \theta_{ m }=\left( m +\frac{1}{2}\right) \frac{\lambda}{a}$
(with secondary maximum) ... (3)
We can imagine the single slit as being made up of a large number of Huygens' sources evenly distributed over the width of the slit. Then the maxima and minima of the pattern arise from the interference of the various Huygens' wavelets.
Now, imagine the single slit as made up of two adjacent slits, each of width $a / 2$. Since, the incident plane wavefronts are parallel to the plane of the slit, all the Huygens sources at the slit will be in phase. They will therefore also in phase at the point $P _0$ on the screen, where $P _0$ is equidistant from all the Huygens sources. At $P _0$, then, we get the central maximum.
For the first minimum of intensity on the screen, the path difference between the waves from the Huygens sources A and $O$ (or $O$ and $B$ ) is $\lambda / 2$, which is the condition for destructive interference. Suppose, the nodal line $O P$ for the first minimum subtends an angle $\theta$ at the slit; $\theta$ is very small. With $P$ as the centre and $P A$ as radius, strike an arc intersecting $P B$ at $C$. Since, $D \gg a$, the arc $A C$ can be considered a straight line at right angles to $P B$. Then, $\triangle A B C$ is a right-angled triangle similar to $\triangle OP P _0 P$.
This means that, $\angle BAC = \theta$
$\therefore BC = a \sin\theta$
$\therefore $ Difference in path length,
$BC = PB – PA = (PB – PO) + (PO – PA)$
$=\frac{\lambda}{2}+\frac{\lambda}{2}$
$=\lambda$ $ \therefore a \sin \theta=\lambda$
$\therefore \sin \theta \simeq \theta=\frac{\lambda}{a} $
( $\because \theta$ is very small and in radian)
The other nodal lines of intensity minima can be understood in a similar way. In general, then, for the with minimum ( $m= \pm 1, \pm 2, \pm 3, \ldots)$.
$\theta_m=\frac{m \lambda}{a}$ (mth minimum) $\ldots$ (2)
as $\theta_m$ is very small and in radian.
Between the successive minima, the intensity rises to secondary maxima when the path difference is an odd-integral multiple of $\frac{\lambda}{2}$ :
$a \sin \theta_m=(2 m+1) \frac{\lambda}{2}=\left(m+\frac{1}{2}\right) \lambda$
i.e., at angles given by,
$\theta_{ m } \simeq \sin \theta_{ m }=\left( m +\frac{1}{2}\right) \frac{\lambda}{a}$
(with secondary maximum) ... (3)
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