Answer:
The distance between the places where the intensity is zero due to the double slit effect is 15 mm.
Explanation:
Given that,
Distance between the slits = 0.04 mm
Width = 0.01 mm
Distance between the slits and screen = 1 m
Wavelength = 600 nm
We need to calculate the distance between the places where the intensity is zero due to the double slit effect
For constructive fringe
First minima from center
[tex]x_{1}=\dfrac{\lambda D}{2d}[/tex]
Second minima from center
[tex]x_{2}=\dfrac{3\lambda D}{2d}[/tex]
The distance between the places where the intensity is zero due to the double slit effect
[tex]\Delta x_{d}=x_{2}-x_{1}[/tex]
[tex]\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}[/tex]
[tex]\Delta x_{d}=\dfrac{\lambda D}{d}[/tex]
Put the value into the formula
[tex]\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}[/tex]
[tex]\Delta x_{d}=0.015 =15\times10^{-3}\ m[/tex]
[tex]\Delta x_{d}=15\ mm[/tex]
Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.
This question involves the concepts of Young's Double Slit Experiment and fringe spacing.
a) Distance Δxa between the places where the intensity is zero due to the single slit effect is "60 mm".
b) Distance Δxd in mm between the places where the intensity is zero due to the double-slit effect is "15 mm".
a)
We will use the formula of Young's double-slit experiment here:
[tex]\Delta x_a = \frac{\lambda L}{d}[/tex]
where,
[tex]\Delta x_a[/tex] = distance between places of zero intensity in case of a single slit
[tex]\Delta x_a[/tex] = fringe spacing in case of a single slit = ?
λ = wavelength = 600 nm = 6 x 10⁻⁷ m
L = distance from screen = 1 m
d = width of a single slit = 0.01 mm = 1 x 10⁻⁵ m
Therefore,
[tex]\Delta x_a=\frac{(6\ x\ 10^{-7}\ m)(1\ m)}{1\ x\ 10^{-5}\ m}\\\\\Delta x_a=6\ x\ 10^{-2}\ m=60\ mm[/tex]
b)
We will use the formula of Young's double-slit experiment here, as well:
[tex]\Delta x_d = \frac{\lambda L}{d}[/tex]
where,
[tex]\Delta x_d[/tex] = distance between places of zero intensity in case of double slits
[tex]\Delta x_d[/tex] = fringe spacing in case of double slits = ?
λ = wavelength = 600 nm = 6 x 10⁻⁷ m
L = distance from screen = 1 m
d = width of a single slit = 0.04 mm = 4 x 10⁻⁵ m
Therefore,
[tex]\Delta x_a=\frac{(6\ x\ 10^{-7}\ m)(1\ m)}{4\ x\ 10^{-5}\ m}\\\\\Delta x_a=1.5\ x\ 10^{-2}\ m=15\ mm[/tex]
Learn more about Young's Double Slit Experiment here:
brainly.com/question/13935986?referrer=searchResults
The attached picture shows Young's Double Slit Experiment.
