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Svetach [21]
3 years ago
12

Light from a helium-neon laser (λ=633nm) passes through a circular aperture and is observed on a screen 4.0 m behind the apertur

e. The width of the central maximum is 2.5 cm .What is the diameter (in mm) of the hole?
Physics
1 answer:
devlian [24]3 years ago
5 0

Answer:

d = 0.247 mm

Explanation:

given,

λ = 633 nm

distance from the hole to the screen = L = 4 m

width of the central maximum = 2.5 cm

                                             2 y = 0.025 m

                                               y = 0.0125 m

For circular aperture

  sin \theta = 1.22\dfrac{\lambda}{d}

using small angle approximation

  \theta = \dfrac{y}{D}

now,

   \dfrac{y}{D} = 1.22\dfrac{\lambda}{d}

   y = 1.22\dfrac{\lambda\ D}{d}

   d = 1.22\dfrac{\lambda\ D}{y}

   d = 1.22\dfrac{633\times 10^{-9}\times 4}{0.0125}

         d =0.247 x 10⁻³ m

         d = 0.247 mm

the diameter of the hole is equal to 0.247 mm

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