Question

Light from a helium-neon laser (λ=633nm) passes through a circular aperture and is observed on a screen 4.0 m behind the aperture. The width of the central maximum is 2.5 cm .What is the diameter (in mm) of the hole?

Answer 1

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