Question

You illuminate a slit with a width of 0.0537 mm with a light of wavelength 719 nm and observe the resulting diffraction pattern on a screen that is situated 2.77 m from the slit. What is the width, in centimeters, of the pattern's central maximum?

Answer 1

Width of central maximum will be
`7.423* 10^(-12)m`

Step-by-step explanation:

We have given width of the slit
`D =0.0537mm=5.37* 10^(-5)m`

Wavelength of light
`\lambda =719nm=719* 10^(-9)m`

Distance of screen from slit D = 2.77 m

Angle to first minimum
`sin\Theta =(\lambda )/(d)=(719* 10^(-19))/(0.0537* 10^(-3))=1.34* 10^(-12)`

So width of the central maximum

w=
`2y=2* Dsin\Theta =2* 2.77* 1.34* 10^(-12)=7.423* 10^(-12)m`

So width of central maximum will be
`7.423* 10^(-12)m`