Question

A single-slit diffraction pattern is formed when light of λ = 759.0 nm is passed through a narrow slit. The pattern is viewed on a screen placed one meter from the slit. What is the width of the slit (mm) if the width of the central maximum is 2.49 cm?

Please show all work!

Answer 1

The width b of central maxima is given by

b = 2Dk/a

Where D is the distrance of screen from slit, k is the wavelength of light and a is width of the slit.

Rearranging the terms, we can write

a= 2Dk/b ...eqn 1

Given, k= 759.0 nm = 759*10^-9 m

D = 1 m

b = 2.49 cm = 0.0249 m

Putting all these values in eqn 1, we get

a= 2*1*759*10^-9/0.0249

= 1518*10^-9/0.0249

= 60963.8554*10^-9 m

= 0.060*10^-3 m

a = 0.060 mm

Which is the required slit width.