
Helium-neon laser light \[(\lambda =632.8\,nm)\] is sent through a \[0.300\] mm wide single slit. What is the width of the central maximum on a screen \[1.00\] m from the slit?
Answer
493.2k+ views
Hint:We know the conditions for constructive and destructive interference by considering two coherent sources of light passing through the two slits and forming fringes on a screen placed at some distance from the slits.
Complete step by step answer:
As we know the conditions for constructive and destructive interferences by using the wave equations for the two light sources and finding the phase difference or the path difference. In the second part of the question, we can use both the wavelengths to find the distance required but the condition is to find the distance of that bright fringe only respective to that wavelength as to which number the bright fringe will coincide in case both wavelengths will be different.
We will find the path difference of the two wavelengths of light and will then find that at what intervals from the central maxima, the constructive or destructive interference will occur. Here from equation we get that $m=1;$ which means our sine will be given by;
$\sin \theta =\dfrac{{{6.32810}^{-7}}m}{3\times {{10}^{-4}}m} \\
\Rightarrow \sin \theta =2.11\times {{10}^{-3}}$
Thus,
$\dfrac{y}{1}=\tan \theta \approx \sin \theta \approx \theta $
where theta is very small.
$ y=2.11\,mm$
$\therefore 2y=2\left( 2.11 \right)=4.22\,mm.$
Hence, the width of the central maximum is 4.22 mm.
Note: Remember that the value of either the momentum or the energy, giving either of the values. In such situations, firstly find the value of the wavelength and then proceed further. The energy and momentum of each photon in the light beam. By making use of the equation for the photon energy and momentum.
Complete step by step answer:
As we know the conditions for constructive and destructive interferences by using the wave equations for the two light sources and finding the phase difference or the path difference. In the second part of the question, we can use both the wavelengths to find the distance required but the condition is to find the distance of that bright fringe only respective to that wavelength as to which number the bright fringe will coincide in case both wavelengths will be different.
We will find the path difference of the two wavelengths of light and will then find that at what intervals from the central maxima, the constructive or destructive interference will occur. Here from equation we get that $m=1;$ which means our sine will be given by;
$\sin \theta =\dfrac{{{6.32810}^{-7}}m}{3\times {{10}^{-4}}m} \\
\Rightarrow \sin \theta =2.11\times {{10}^{-3}}$
Thus,
$\dfrac{y}{1}=\tan \theta \approx \sin \theta \approx \theta $
where theta is very small.
$ y=2.11\,mm$
$\therefore 2y=2\left( 2.11 \right)=4.22\,mm.$
Hence, the width of the central maximum is 4.22 mm.
Note: Remember that the value of either the momentum or the energy, giving either of the values. In such situations, firstly find the value of the wavelength and then proceed further. The energy and momentum of each photon in the light beam. By making use of the equation for the photon energy and momentum.
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