Question

A single slit 1.4 mmmm wide is illuminated by 460-nmnm light. Part A What is the width of the central maximum (in cmcm ) in the diffraction pattern on a screen 5.0 mm away

Answer 1

1.643*10⁻⁴cm

Step-by-step explanation:

In a single slit experiment, the distance on a screen from the centre point is expressed as y =
`(\delta m \lambda d)/(a)` where;

`\delta m` is the first two diffraction minima = 1

`\lambda` is light wavelength

d is the distance of diffraction pattern from the screen

a is the width of the slit

Given
`\lambda` = 460-nm = 460*10⁻⁹m

d = 5.0mm = 5*10⁻³m

a = 1.4mm = 1.4*10⁻³m

Substituting this values into the formula above to get width of the central maximum y;

y = 1*460*10⁻⁹ * 5*10⁻³/1.4*10⁻³

y = 2300*10⁻¹²/1.4*10⁻³

y = 1642.86*10⁻⁹

y = 1.643*10⁻⁶m

Converting the final value to cm,

since 100cm = 1m

x = 1.643*10⁻⁶m

x = 1.643*10⁻⁶ * 100

x = 1.643*10⁻⁴cm

Hence, the width of the central maximum in the diffraction pattern on a screen 5.0 mm away is 1.643*10⁻⁴cm