Answer:
the wavelength of the light used is 4.8 × 10⁻⁷m
Explanation:
We have given distance between the two slits
[tex]d=1.6\times 10^{-2}mm=1.6\times 10^{-5}m[/tex]
Angle [tex]\Theta =8.8^{\circ}[/tex]
We know that for nth order fringe
[tex]dsin\Theta =n\lambda[/tex]
In question it is given that 5th order so n =5
[tex]So 1.6\times 10^{-5}sin\ 8.8 =5\lambda\\\lambda = 4.8\times 10^{-7}m[/tex]
Given Information:
distance = d = 0.016 mm = 0.0016 cm = 0.000016 m
Angle = θ = 8.8°
bright spot = m = 5
Required Information:
Wavelength = λ = ?
Answer:
Wavelength = 4.89x10⁻⁷ m
Explanation:
Monochromatic light passes through a double slit. The corresponding diffraction is given by the equation:
dsinθ = mλ
where d is the distance between two slits, m is the order of the diffraction, θ is the angle and λ is the wavelength of the light wave.
λ = dsinθ/m
λ = 0.016*sin(8.8°)/5
λ = 4.89x10⁻⁷ m
Therefore, the wavelength of the monochromatic light falling on two slits is 4.89x10⁻⁷ m
