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You illuminate a slit with a width of 0.0537 mm with a light of wavelength 719 nm and observe the resulting diffraction pattern on a screen that is situated 2.77 m from the slit. What is the width, in centimeters, of the pattern's central maximum?

Respuesta :

Answer:

Width of central maximum will be [tex]7.423\times 10^{-12}m[/tex]

Explanation:

We have given width of the slit [tex]D =0.0537mm=5.37\times 10^{-5}m[/tex]

Wavelength of light [tex]\lambda =719nm=719\times 10^{-9}m[/tex]

Distance of screen from slit D = 2.77 m

Angle to first minimum [tex]sin\Theta =\frac{\lambda }{d}=\frac{719\times 10^{-19}}{0.0537\times 10^{-3}}=1.34\times 10^{-12}[/tex]

So width of the central maximum

w= [tex]2y=2\times Dsin\Theta =2\times 2.77\times 1.34\times 10^{-12}=7.423\times 10^{-12}m[/tex]

So width of central maximum will be [tex]7.423\times 10^{-12}m[/tex]

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How much money is 16 2/3 of You illuminate a slit with a width of 00537 mm with a light of wavelength 719 nm and observe the resulting diffraction pattern on a screen that is situated 277

Respuesta :

Answer:

Width of central maximum will be [tex]7.423\times 10^{-12}m[/tex]

Explanation:

We have given width of the slit [tex]D =0.0537mm=5.37\times 10^{-5}m[/tex]

Wavelength of light [tex]\lambda =719nm=719\times 10^{-9}m[/tex]

Distance of screen from slit D = 2.77 m

Angle to first minimum [tex]sin\Theta =\frac{\lambda }{d}=\frac{719\times 10^{-19}}{0.0537\times 10^{-3}}=1.34\times 10^{-12}[/tex]

So width of the central maximum

w= [tex]2y=2\times Dsin\Theta =2\times 2.77\times 1.34\times 10^{-12}=7.423\times 10^{-12}m[/tex]

So width of central maximum will be [tex]7.423\times 10^{-12}m[/tex]