From Young's single slit experiment, the distance away from the wall will be 1.068 m
Given that 587.9 nm of wavelength of light passes through a single slit 0.73 mm wide, it creates a diffraction pattern.
From the question, the following parameters are given:
The wavelength of the light λ = 587.9 nm
The width of the slit a = 0.73 mm
Fringe width X = 0.86 mm
The distance away from the wall D = ?
The fringe width is related to the wavelength of the light source by the equation:
X = Dλ ÷ a
Substitute all the parameters into the formula
0.83 × [tex]10^{-3}[/tex] = 587.9 × [tex]10^{-9}[/tex] D ÷ 0.73 ×
Cross multiply
587.9 × [tex]10^{-9}[/tex] D = 6.278 × [tex]10^{-7}[/tex]
make D the subject of the formula
D = 6.278 × [tex]10^{-7}[/tex] ÷ 587.9 × [tex]10^{-9}[/tex]
D = 1.068 m
Therefore, the distance away from the wall is 1.068 m
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If the first minimum is [tex]0.86 mm[/tex] from the central maximum, the distance away is 1.07 meters.
Given the data in the question;
From Thomas Young's single slit experiment:
[tex]\frac{a*y}{L} = m * \lambda[/tex]
Where a is the width of the slit, y is first minimum, L is the distance, m is the order number and λ is the wavelength.
We substitute our values into the equation
[tex]\frac{0.00073m\ *\ 0.00086m}{L} = 1\ *\ ( 5.879*10^{-7}m)\\\\\frac{0.0000006278m^2}{L} = 5.879*10^{-7}m\\\\L = \frac{0.0000006278m^2}{5.879*10^{-7}m} \\\\L = 1.07m[/tex]
Therefore, if the first minimum is [tex]0.86 mm[/tex] from the central maximum, the distance away is 1.07 meters.
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