Question

Light from a He-Ne laser of wavelength 633 nm passes through a circular aperture. It is observed on a screen 4.0 m behind the aperture. The width of the central maximum is 1.1 cm. What is the diameter of the hole

Answer 1

The diameter of the hole is 5.6164 μm

Step-by-step explanation:

Given the data in the question;

wavelength λ = 633 nm = 633 × 10⁻⁹ m

L = 4.0 M

width of the central maximum D = 1.1 cm.

Now,

Y = D / 2

Y = 1.1 / 2 = 0.55 cm

So diameter of the aperture will be;

α = 1.22λ / ( Y / L )

we substitute in our values;

α = ( 1.22 × (633 × 10⁻⁹ m)) / ( 0.55 / 4.0 )

α = 0.00000077226 / 0.1375

α = 5.6164 × 10⁻⁶ m

α = ( 5.6164 × 10⁻⁶ × 10⁶ ) μm

α = 5.6164 μm

Therefore, the diameter of the hole is 5.6164 μm