Fraunhofer Diffraction By A Single Slit - Practice Questions & MCQ

Fraunhofer diffraction by a single slit

let's assume  a plane wave front is incident on a slit AB (of width b). Each and every part of the exposed part of the plane wave front (i.e. every part of the slit) acts as a source of secondary wavelets spreading in all directions. The diffraction is obtained on a screen placed at a large distance. (In practice, this condition is achieved by placing the screen at the focal plane of a converging lens placed just after the slit).

•  The diffraction pattern consists of a central bright fringe (central maxima) surrounded by dark and bright lines (called secondary minima and maxima).
•  At point O on the screen, the central maxima is obtained. The wavelets originating from points A and B meets in the same phase at this point, hence at O, intensity is maximum

 Secondary minima : For obtaining nth secondary minima at P on the screen, path difference between the diffracted waves 

$
\Delta x=b \sin \theta=n \lambda
$

1. Angular position of nth secondary minima:

$
\sin \theta \approx \theta=\frac{n \lambda}{b}
$

2. Distance of nth secondary minima from central maxima:

$
\begin{aligned}
x_n=D \cdot \theta=\frac{n \lambda D}{b} \quad & \\
& \\
& \\
& \approx D=D=\text { Focal length of converging lens. }
\end{aligned}
$

Secondary maxima: For nth secondary maxima at P on the screen.
Path difference

$
\Delta x=b \sin \theta=(2 n+1) \frac{\lambda}{2} ; \text { where } \mathrm{n}=1,2,3 \ldots \ldots
$

(i) Angular position of nth secondary maxima

$
\sin \theta \approx \theta \approx \frac{(2 n+1) \lambda}{2 b}
$

(ii) Distance of nth secondary maxima from central maxima:

$
x_n=D \cdot \theta=\frac{(2 n+1) \lambda D}{2 b}
$

Central maxima : The central maxima lies between the first minima on both sides.
(i) The Angular width d central maxima $=2 \theta=\frac{2 \lambda}{b}$
(ii) Linear width of central maxima

$
=2 x=2 D \theta=2 f \theta=\frac{2 \lambda f}{b}
$
 

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