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

83.55 μm

Step-by-step 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 =
`(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 = (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