In double slit experiment, the two slits are 1 mm apart and the screen is placed 1m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining 10 Maxima of double slit, within the central Maxima of Single Split pattern?
Question
In double slit experiment, the two slits are 1 mm apart and the screen is placed 1m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining 10 Maxima of double slit, within the central Maxima of Single Split pattern?
Solution
The double slit experiment is a demonstration that light and other forms of electromagnetic radiation can also exhibit behavior typical of particles, as when they are small enough, such as electrons. This is known as wave-particle duality.
The formula for the position of the maxima in a double slit experiment is given by:
dsin(θ) = mλ
where:
- d is the distance between the slits,
- θ is the angle between the central maximum and the mth maximum,
- m is the order of the maximum, and
- λ is the wavelength of the light.
The condition for obtaining 10 maxima of double slit within the central maxima of single slit pattern is that the 10th maximum of double slit pattern should coincide with the first minimum of single slit pattern.
The formula for the position of the minima in a single slit experiment is given by:
asin(θ) = mλ
where:
- a is the width of the slit,
- θ is the angle between the central maximum and the mth minimum,
- m is the order of the minimum, and
- λ is the wavelength of the light.
For the 10th maximum of double slit pattern to coincide with the first minimum of single slit pattern, we equate the two formulas:
dsin(θ) = 10λ = a*sin(θ)
Solving for a, we get:
a = d*λ/10
Substituting the given values:
a = (1 mm)*(500 nm)/10 = 50 nm
So, the width of each slit for obtaining 10 maxima of double slit within the central maxima of single slit pattern is 50 nm.
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