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 × = 587.9 × D ÷ 0.73 ×
Cross multiply
587.9 × D = 6.278 ×
make D the subject of the formula
D = 6.278 × ÷ 587.9 ×
D = 1.068 m
Therefore, the distance away from the wall is 1.068 m
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