Answer
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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.
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.
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