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

The diameter of the hole is approximately 0.0201 mm.

Step-by-step explanation:

The width of the central maximum in a diffraction pattern can be determined using the formula:

w = (2 * λ * D) / x

Where w is the width of the central maximum, λ is the wavelength of the light, D is the distance between the aperture and the screen, and x is the diameter of the hole. Rearranging the formula, we can solve for x:

x = (2 * λ * D) / w

Plugging in the given values, we get:

x = (2 * 633 * 10^-9 * 4.0) / 0.025

x ≈ 0.0201 mm

Answer 2

d = 0.247 mm

Step-by-step 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(\lambda)/(d)`

using small angle approximation

`\theta = (y)/(D)`

now,

`(y)/(D) = 1.22(\lambda)/(d)`

`y = 1.22(\lambda\ D)/(d)`

`d = 1.22(\lambda\ D)/(y)`

`d = 1.22(633* 10^(-9)* 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