Question

Calculate the wavelength of light that has its third minimum at an angle of 30.0º when falling on double slits separated by 3.00~\mu\text{m}3.00 μm.

Answer 1

Answer:

λ = 428.57 nm

Explanation:

Young's double slit experiment is described by the expression

          d sin θ = m λ                    constructive interference

          d sin θ = (m + ½) λ           destructive interference

where d is the distance between the slits, λ is the wavelength, m is an integer that describes the order of interference

In this case they give us the angle 30º, the distance between the slits d = 3.00 μm = 3.00 10⁻⁶ m and the order of interference m = 3; we use the destructive interference equation

       

         λ = d sin θ / (m + ½)

let's calculate

        λ = 3 10⁻⁶ sin 30 / (3 +1/2)

         

         λ = 4.2857 10⁻⁷ m

we reduce to nm

        λ = 4.2857 10⁻⁷ (10⁹ nm / 1 m)

         λ = 428.57 nm

Answer 2

The wavelength of light will be "428.57 nm".

Given values:

• Angle = 30.0º
• Order of interference (m) = 3
• Distance b\w the slits (d) = 3.00 μm or $3\times 10^{-6}$

As we know,

→ $\lambda = \frac{d \ Sin \theta}{(m+\frac{1}{2} )}$

By putting the values, we get

      $= \frac{3\times 10^{-6} Sin 30^{\circ}}{(3+\frac{1}{2} )}$

      $= 4.2857\times 10^{-7} \ m$

By converting it to "nm", we get

      $= 4.2857\times 10^{-7}\times \frac{10^9 \ nm}{1 \ m}$

      $= 428.57 \ nm$

Thus the response above is correct.  

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