Originally Posted by **ling233** For a double slit experiemnt, the bright fringe has intensity of I, when one of the slit is covered, the intensity is 0.25I.
In my opinion, the slit now become single slit, so it has area of half of the initial area, but we know that power is the product of intensity x area, by keeping the power constant. So it should has intensity of 2I if one of the slit is covered? |

Light is a transverse wave having electric (E) and magnetic (B) fields.

The intensity is proportional to the square of the electric field, i. e. E^2 is proportionalto I.

If you have two slits as in the Young´s double slit experiment you will have

E_total_two_slits = (E+E)

(E_total_two_slits)^2 = 4E^2

I_two_slits is proportional to 4E^2

Let us say one of the slits is closed. In this case, you have

E_total_one_slit = E

(E_total_one_slit)^2 = E^2

I_one_slit is proportional to E^2

As you see the intensity gets decrease to one fourth of the initial case.