Question

Monochromatic light with wavelength 590 nm passes through a single slit 2. 30 ?m wide and 1. 90 m from a screen. Find the distance between the first- and second-order dark fringes on the screen.

Answer 1

The distance between the first and second-order dark fringe is 0.441 m.

Diffraction of a single slit:

When the light wave passes through a single slit of width which is comparable to the wavelength of the light, then the light wave bends at the edges of the slit. This is called diffraction.

Note: It is assumed that the slit is 3*10^(-3) mm wide. 1 nm = 10^(-9) m and 1mm = 10^(-3) m.

The dark fringes are obtained at the position which satisfies the equation,

d*sinθ = mλ

where d is the slit width, λ is the wavelength of the wave, θ is the angle of diffraction and m denotes the order of the dark fringe.

For first order fringe (m=1), the angle of diffraction θ₁ is,

d*sinθ₁ = λ

sinθ₁ =λ/d

Substitute λ=590 nm, d=3*10^(-3) mm, and solve it.

sinθ₁ =(590 nm)/(3*10^(-3) mm)

sinθ₁ =(590*10^(-9))/ (3*10^(-3)*10^(-3) m)

sinθ₁ =0.196

θ₁ =11.03 degree

Similarly, for second-order fringe (m=2), the angle of diffraction θ₂ is,

d*sinθ₂= 2λ

sinθ₂ =2λ/d

Substitute λ=590 nm, d=3*10^(-3) mm, and solve it.

sinθ₂ =(2*590 nm)/(3*10^(-3) mm)

sinθ₂ =(2*590*10^(-9))/ (3*10^(-3)*10^(-3) m)

sinθ₂ =0.393

θ₂ =23.14 degree

From geometry, the positions x₁ and x₂ of the first and second-order dark fringe from the center of the screen are x₁=Dtanθ₁ and x₂= Dtanθ₂ where D is the distance of the screen from the slit. The distance s between the first-order and second-order dark fringe is then given by,

s=D(tanθ₂-tanθ₁)

Substitute D=1.90 m, θ₁=11.03 degree, and θ₂=23.14 degree in this equation and solve it.

s=1.90*(tan(23.14)-tan(11.03))

s=1.90*(0.427-0.195)

s=0.441 m

Learn more about diffraction here:

brainly.com/question/12290582

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