Answer:
The position of the first dark spot on the positive side of the central maximum is 1.26 mm.
Explanation:
Given that,
Wavelength of light is 633 nm.
Slit width, d = 0.5 mm
The diffraction pattern forms on a screen 1 m away from the slit. We need to find the position of the first dark spot on the positive side of the central maximum.
For destructive interference :
[tex]\dfrac{dY}{D}=n\lambda[/tex]
Y is the distance of the minima from central maximum
Here, n = 1
[tex]Y=\dfrac{n\lambda D}{d}\\\\Y=\dfrac{1\times 633\times 10^{-9}\times 1}{0.5\times 10^{-3}}\\\\Y=0.00126\ m\\\\Y=1.26\times 10^{-3}\ m\\\\Y=1.26\ mm[/tex]
So, the position of the first dark spot on the positive side of the central maximum is 1.26 mm.
This question involves the concept of Young's Double Slit Experiment formula and fringe spacing.
The position of the first dark spot on the positive side of the central maximum is "1.27 mm".
Young's double-slit formula can be written as follows:
[tex]\Delta x=\frac{\lambda L}{d}[/tex]
where,
Δx = fringe spacing = distance of first dark spot from central maximum = ?
λ = wavelength of light = 633 nm = 6.33 x 10⁻⁷ m
L = screen distance = 1 m
d = slit width = 0.5 mm = 5 x 10⁻⁴ m
Therefore,
[tex]\Delta x = \frac{(6.33\ x\ 10^{-7}\ m)(1\ m)}{5\ x\ 10^{-4}\ m}[/tex]
Δx = 1.27 x 10⁻³ m = 1.27 mm
Learn more about Young's Double Slit Experiment here:
https://brainly.com/question/13935986?referrer=searchResults
The attached picture shows Young's Double Slit Experiment.
