Question

Light of wavelength 618.0 nm 618.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 84.5 cm 84.5 cm from the slit. The distance on the screen between the second order minimum and the central maximum is 1.25 cm 1.25 cm . What is the width a a of the slit in micrometers (μm)?

Answer 1

Answer:

83.55 μm

Explanation:

In a single slit experiment, the distance (on the screen), y, between the central maximum and the minimum of order, m, is given as:

y = $\frac{md}{a}$ * λ

where λ = wavelength

a = width of slit

d = distance between slit and screen

To get the width of the slit, we make a the subject of formula:

a = (mλd)/y

$a = \frac{2 * 618 * 10^{-9} * 84.5*10^{-2}}{1.25 * 10^{-2}}$

(All units of length are in meters)

$a = 8.355 * 10^{-5} m\\\\\\a = 83.55 * 10^{-6} m$

a = 83.55 μm