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Fraunhofer Diffraction By A Single Slit - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Syllabus

Quick Facts

Fraunhofer diffraction by a single slit is considered one the most difficult concept.
20 Questions around this concept.

Solve by difficulty

EASY

The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young’s doubleslit experiment is

infinite

five

three

zero

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Concepts Covered - 1

Fraunhofer diffraction by a single slit

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

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

Angular position of nth secondary minima:

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

2. Distance of nth secondary minima from central maxima:

$x_{n}=D\cdot \theta =\frac{n\lambda D}{b}$ where D = Distance between slit and screen.

$f\approx D$ = Focal length of converging lens.

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

Path difference $\Delta x =b\sin{\theta }=(2n+1) \frac{\lambda}{2}$ ; where n = 1, 2, 3 .....

(i) Angular position of nth secondary maxima

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

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

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

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

Intensity distribution: if the intensity of the central maxima is $I_{0}$ then the intensity of the first and secondary maxima are
found to be $\frac{I_{0}}{22}$ and $\frac{I_{0}}{61}$ . Thus diffraction fringes are of unequal width and unequal intenstities.

(i) The mathematical expression for in intensity distribution on the screen is given by:

$I=I_{o}\left(\frac{\sin \alpha}{\alpha}\right)^{2}$ where $\alpha$ is just a convenient connection between the angle $\theta$ that locates a point on the viewing screening and light intensity I.

$\phi=$ Phase difference between the top and bottom ray from the slit width b.
Also $\alpha=\frac{1}{2} \phi=\frac{\pi b}{\lambda} \sin \theta$ .

(ii) As the slit width increases relative to wavelength the width of the control diffraction maxima decreases; that is, the light undergoes less flaring by the slit. The secondary maxima also decreases in width and becomes weaker.

(iii) If b>>λ, the secondary maxima due to the slit disappear; we then no longer have single slit diffraction.