Question

In Fraunhofer diffraction pattern due to a single slit, the slit of width $0.1mm$ is illuminated by monochromatic light of wavelength $600nm$ . 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) $0.1m$
(C) $6m$
(D) $100m$

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;
 $\alpha ={\displaystyle \frac{2\lambda }{W}}$
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, $\lambda =600nm$ .
The width of the single slit is, $W=0.1mm$ .
Fraunhofer diffraction pattern due to a single slit formula is given as;
 $\alpha ={\displaystyle \frac{2\lambda }{W}}$
Substitute the values of wavelength of the monochromatic light source and the width of the single slit in the above Fraunhofer diffraction formula;
 $\alpha ={\displaystyle \frac{2\times \left(600\times {10}^{-9}m\right)}{0.1\times {10}^{-3}m}}$
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.