A 0.51mm -diameter hole is illuminated by light of wavelength 510nm . What is the width of the central maximum on a screen 2.0m behind the slit? Answer in delta y=

Respuesta :

delta y = lamda L/d

you will find that the width of the central maximum on a screen 2.0 m behind the slit would be :
1/30E-12

hope this helps

Answer:

 y = 2.44 10⁻³ m = 2.44 mm

Explanation:

The diffraction phenomenon is described by the expression

         a sin θ = m λ

Where a is the width of the slit, λ the wavelength and m the diffraction power, to have the first zero m must be one (m = 1) so this distance gives the width of the central maximum torque m = 0 up to the first minimum.

In the case of circular aperture we must use polar coordinates, also as in these experiments the angle is very small we can approximate the sine by its angle

        tan θ = y / x

        tan θ = sin θ / cos θ = sin θ

        sin θ = y / x

       

The use of polar coordinates introduces a numerical constant, therefore the expression remains

           a  y / x = 1.22 λ

          y = 1.22 λ x / a

Let's calculate

          y = 1.22  510 10⁻⁹ 2.0 / 0.51 10⁻³

          y = 2.44 10⁻³ m = 2.44 mm

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