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In a two slit interference experiment, what is the separation between the two slits for which 610-nm orange light has its first maximum at an angle of 30.0º?

Respuesta :

Answer: d = [tex]1.22 *10^{-6}m[/tex]

Step-by-step explanation: for  a double slit interference experiment, the formulae relating parameters is

d*sinθ = mλ

d = distance between slits

m = integral order of pattern=1

θ = angle between light formed from slit to screen = 30

λ  = wavelength of light= 610nm= [tex]6*10^{-7} m[/tex]

hence d = mλ /sinθ

d = [tex]\frac{1*6.10*10^{-7} }{sin 30}[/tex]

d = [tex]\frac{1*6.10*10^{-7} }{0.5}[/tex]

d = [tex]1.22* 10^{-6} m[/tex]

fichoh

The seperation between the two slits, d is related to the wavelength, and angle using the formula below ; Hence, the value of d is [tex]1.22 \times 10^{-6} \: meters [/tex]

Using the relation :

  • [tex] d = \frac{mλ}{sin\theta}[/tex]

  • d = distance between the slits

  • θ = 30°

  • λ = wavelength of light = 610 nm = [tex]6.1 \times10^{-7}[/tex]

  • m = 1

Substituting the values into the equation :

[tex] d = \frac{1 \times 6.1 \times 10^{-7}}{0.5} = 0.00000122 = 1.22 \times 10^{-6} \:meteres[/tex]

Therefore, the seperation between the two slits is [tex]1.22 \times 10^{-6} \: meters [/tex]

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