Answer:
69.7 cm
Explanation:
What is the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe?
For a diffraction grating, dsinθ = mλ where d = grating spacing = 1/(3000 lines per cm) = 1/3000 × 100 m per line = 1/300000 m = 1/3 × 10⁻⁵ m, m = order of fringe and λ = wavelength of light = 550 nm = 550 × 10⁻⁹ m.
Also, tanθ = x/D where x = distance of nth order fringe from central maximum and D = distance of screen from grating = 115 cm = 1.15 m
Now sinθ = d/mλ, Since θ is small, sinθ ≅ tanθ
So, d/mλ = x/D for a second order bright fringe, m = 2.
So, d/2λ = x/D
x = dD/2λ
So, x =
For a dark fringe, we have
d/(m + 1/2)λ = x'/D where x' is the distance of the fringe from the central maximum.
For a second-order dark fringe, m = 2. So,
d/(2 + 1/2)λ = x'/D
d/(5/2)λ = x'/D
2d/5λ = x'/D
x' = 2dD/5λ
So, the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe is x" = dD/2λ - 2dD/5λ
x" = dD/10λ
Substituting the values of the variables into the equation, we have
x"= 1/3 × 10⁻⁵ m × 1.15 m/(10 × 550 × 10⁻⁹ m)
x" = 1.15/165 × 10² m
x" = 0.00697 × 10² m
x" = 0.697 m
x" = 69.7 cm
The distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe 69.7 cm
Diffraction of light occurs when a light wave passes by a corner or through an opening or slit that is physically the approximate size of, or even smaller than that light's wavelength
For a diffraction grating, dsinθ = mλ
where,
d = grating spacing = 1/(3000 lines per cm) = 1/3000 × 100 m per line = 1/300000 m = 1/3 × 10⁻⁵ m,
m = order of fringe
λ = wavelength of light = 550 nm = 550 × 10⁻⁹ m.
Also, [tex]tan\theta=\dfrac{x}{D}[/tex]t where
x = distance of nth order fringe from central maximum
D = distance of screen from grating = 115 cm = 1.15 m
Now [tex]Sin\theta =\dfrac{d}{m\lambda}[/tex] , Since θ is small, sinθ ≅ tanθ
So, [tex]\dfrac{d}{m\lambda}=\dfrac{x}{D}[/tex]
for a second order bright fringe, m = 2.
So, [tex]\dfrac{d}{2\lambda}=\dfrac{x}{D}[/tex]
[tex]x=\dfrac{dD}{2\lambda}[/tex]
For a dark fringe, we have
[tex]\dfrac{d}{(m+\dfrac{1}{2})\lambda}=\dfrac{X'}{D}[/tex]
where x' is the distance of the fringe from the central maximum.
For a second-order dark fringe, m = 2. So,
[tex]\dfrac{d}{(m+\dfrac{1}{2})\lambda}=\dfrac{X'}{D}[/tex]
[tex]\dfrac{d}{ \dfrac{5}{2}\lambda}=\dfrac{X'}{D}[/tex]
[tex]X'=\dfrac{2dD}{5\lambda}[/tex]
So, the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe is
[tex]X''=\dfrac{dD}{2\lambda}-\dfrac{2dD}{5\lambda}[/tex]
[tex]X''=\dfrac{dD}{10\lambda}[/tex]
Substituting the values of the variables into the equation, we have
[tex]x''=\dfrac{1}{3\times10^{-5}}\times \dfrac{1.15}{10\times 550\times 10^{-9}}[/tex]x
x" = 0.00697 × 10² m
x" = 0.697 m
x" = 69.7 cm
Hence the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe 69.7 cm
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