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550 nm light passes through a diffraction grating with 3000 lines per centimeter. The screen is 115 cm away from the grating. What is the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe?

Sagot :

oladayoademosu

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

shubhamchouhanvrVT

The distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe 69.7 cm

What is difraction?

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

To know more about Diffraction follow

https://brainly.com/question/16749356

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