Question

A 0.50-mm-diameter hole is illuminated by light of wavelength 500 nm. What is the width of the central maximum on a screen 2.0 m behind the slit?

Answer 1

Answer:

$w=4.88*10^{-3}m\\or\\w=4.88mm$

Explanation:

Given data

Diameter D=0.50 mm

Wavelength λ=500 nm

Length L=2.0 m

To find

Width w

Solution

Circular aperture of diameter D will have bright central maximum of diameter:

w=(2.44λL)/D

$w=(\frac{2.44(500*10^{-9}m )(2.0m)}{0.50*10^{-3}m} )\\w=4.88*10^{-3}m\\or\\w=4.88mm$