Say you have a slit with light passing through it. You will get a diffraction pattern, lets call it E ⃗ s E ⃗ s . If you cover the slit with a piece of material that fits just inside the slit, then there is no E ⃗ E ⃗ field in the Fraunhofer limit. The is means that the E ⃗ E ⃗ field of the blocker, lets call it v e c E b v e c E b , must exactly cancel E ⃗ s E ⃗ s . The only way this can happen is if E ⃗ b = − E ⃗ s E ⃗ b = − E ⃗ s . Now if you take the slit away the E ⃗ E ⃗ field of the blocker must still remain and then irradiance must be I b = ( − E s ) 2 = I s I b = ( − E s ) 2 = I s The interference pattern looks the same. You can verify this yourself by taking a strand of your hair and a laser pointer. Shine the laser pointer at a wall and then put a strand of your hair in front of the light beam. The resulting interference pattern is the exact same as one would obtain from a slit with the same width as your hair.

We can use Babinet's Principle to solve complex problems. For example, say you have square aperture with sides of length L L . The the diffraction pattern for light passing through it is E ⃗ = ɛ ⃗ A e i ( k r − ω t ) L 2 R [ sin β L β L ] [ sin α L α L ] E ⃗ = ɛ ⃗ A e i ( k r − ω t ) L 2 R [ sin β L β L ] [ sin α L α L ] where (assuming the aperture lies in the y − z y − z plane) β L = k L Y / 2 R β L = k L Y / 2 R α = k L Z / 2 R . α = k L Z / 2 R . Now put an opaque square of length d d in the middle of the aperture. Now the resulting E ⃗ E ⃗ field is E ⃗ = ɛ ⃗ A R e i ( k r − ω t ) [ L 2 sin β L β L sin α L α L − d 2 sin β d β d sin α d α d ] E ⃗ = ɛ ⃗ A R e i ( k r − ω t ) [ L 2 sin β L β L sin α L α L − d 2 sin β d β d sin α d α d ] or I = I 0 [ L 2 sin β L β L sin α L α L − d 2 sin β d β d sin α d α d ] 2 I = I 0 [ L 2 sin β L β L sin α L α L − d 2 sin β d β d sin α d α d ] 2 β L = k L Y / 2 R β L = k L Y / 2 R α L = k L Z / 2 R . α L = k L Z / 2 R . β d = k d Y / 2 R β d = k d Y / 2 R α d = k d Z / 2 R . α d = k d Z / 2 R .