Question

In Fraunhofer diffraction pattern due to a single slit, the slit of width $0.1\,\,mm$ is illuminated by monochromatic light of wavelength $600\,\,nm$ . What is the ratio of separation between the central maximum and first secondary minimum to the distance between screen and the slit?
(A) $6 \times {10^{ - 3}}\,\,m$
(B) $0.1\,\,m$
(C) $6\,\,m$
(D) $100\,\,m$

Answer 1

Hint
In the field of optics, the Fraunhofer diffraction equation is used to shape the diffraction of waves when the diffraction pattern is observed at a long distance from the diffracting object, and also when it is observed at the focal plane of an imaging lens.
Fraunhofer diffraction formula is given as;
$\alpha = \dfrac{{2\lambda }}{W}$
Where, $\alpha$ denotes the Fraunhofer diffraction, $\lambda$ denotes the wavelength of the monochromatic light source, $W$ denotes the width of the single slit.

Complete step by step answer
The data’s that are given in the problem are;
The wavelength of the monochromatic light source, $\lambda = 600\,\,nm$ .
The width of the single slit is, $W = 0.1\,\,mm$ .
Fraunhofer diffraction pattern due to a single slit formula is given as;
$\alpha = \dfrac{{2\lambda }}{W}$
Substitute the values of wavelength of the monochromatic light source and the width of the single slit in the above Fraunhofer diffraction formula;
$\alpha = \dfrac{{2 \times \left( {600 \times {{10}^{ - 9}}\,\,m} \right)}}{{0.1 \times {{10}^{ - 3}}\,\,m}}$
Changing the notations into meter for easier calculation;
$\alpha = 6 \times {10^{ - 3}}\,\,m$
Therefore, separation between the central maximum and first secondary minimum to the distance between screen and the is given as $\alpha = 6 \times {10^{ - 3}}\,\,m$ .
Hence, the option (A) $\alpha = 6 \times {10^{ - 3}}\,\,m$ is the correct answer.

Note
In the double-slit experiment it is an illustration that light and matter can exhibit attributes of both traditionally explained waves and particles; furthermore, it exhibits the radically prospect nature of quantum mechanical phenomena.