Monochromatic light with lambda equals 630 space n m is incident on a single slit. The slit width is 0.8 mm. If the distance between the screen and slit is 1.8 m, what is the width of the central bright fringe on the screen

The angular width is "" and the linear width is "".

Explanation:

Given:

Wavelength,

or,

Slit width,

or,

Distance between slit and screen,

As we know,

The central bright fringe's angular height,

⇒

and,

The linear width will be:

⇒

By substituting the values, we get

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures