Question

A slit of width 0.12 mm is illuminated with monochromatic light of wavelength 465.2 nm, and a diffraction pattern is formed on a screen 1.54 m away from the slit. What is the width of the central maximum in millimeters?

Answer 1

Through some dedcution, we can find out that the width of the central maximum can be written as:

`\Delta x=2((\lambda D)/(d))`

Thus, for our case:

`\Delta x=2*((465.2*10^(-9)*1.54)/(0.12*10^(-3)))=0.01194m=11.94mm`

Thus, the width of the central maximum is 11.94mm