A student performing a double-slit experiment is using a green laser with a wavelength of 590 nm. She is confused when the m = 5 maximum does not appear. She had predicted that this bright fringe would be 1.6 cm from the central maximum on a screen 1.5 m behind the slits. a. What is the width of her slits? b. Explain what prevented the fth maximum from being observed.?

a) the number of visible stripes is limited by the first zero of the diffraction pattern a = 0.554 10⁻⁴ m, b) d = 2.77 10⁻⁴ m

Explanation:

b) The double slit experiment is described by the equation

d sin θ = m λ

Let's use trigonometry against the breast

tan θ = y / L

tan θ = sin θ / cos θ

For this experiments the angle is small, so we can approximate the tangent

tan θ = sin θ

d y / L = m λ

d = m λ L / y

d = 5 590 10⁻⁹ 1.5 / 0.016

d = 2.77 10⁻⁴ m

a) In the slit experiments the diffraction and interference phenomena always go together. In general, the interference phenomenon gives a pattern of stripes of equal intensity, but the diffraction phenomenon of each slit gives a pattern that has a wide central maximum and some much smaller secondary maximums.

When we observe the phenomenon we see the sum of the two, so the number of visible stripes is limited by the first zero of the diffraction pattern depending on the width of each slit.

If we write the equations of the two phenomena and the divided

d sin θ / a sin θ = m λ / n λ

Where m and n are integers for interference and diffraction phenomena, respectively. For the first cero of diffractio n = 1

d / a = m

a = d / m

a = 2.77 10⁻⁴ / 5

a = 0.554 10⁻⁴ m

This is the width of each slit