The double slit experiment demonstrates the interference patterns of light waves passing through two slits. Light rays passing through the slits accumulate path differences and can undergo constructive or destructive interference when reaching the detection screen. Constructive interference occurs when the path difference is an integer multiple of the wavelength, appearing as bright fringes. Destructive interference happens when the path difference is an odd integer plus a half integer multiple of the wavelength, appearing as dark fringes. For an example using a red light of 620nm wavelength through slits 1.00e-5m apart, the angle to see the second bright fringe is calculated to be 7.12 degrees.

1. Double Slit Experiment
The double slit experiment demonstrates interference effects of light waves. The geometry of
light rays passing through two slits is shown below. The rays, separated by a distance d, strike a
screen at a distance L apart.
The path difference travelled by the two rays can be approximated by: dsin
as long as Fraunhofer's condition is met, which states that the distance L has to be much greater
than the slit spacing d. in the equation equals the deflection angle to a certain point on the
screen, which also equals the upper angle in the right angled triangle.
For the rays to form constructive interference when they are in phase (corresponding to the
bright regions on the screen) then:
dsin = m ,m = 1, 2,...
Also, for destructive interference detected at the screen as dark bands, the condition corresponds
to the path difference as equal to an odd number of half wavelengths:
dsin = (m + ) ,m = 1, 2,...
Note: the bands formed on the screen are referred to bright (constructive interference) and dark
(destructive interference) fringes.

2. Example: If a red light with a wavelength of 620 nm shines on a double slit with the slits cut
1.00e-5 m apart, determine what angle you should look away from the central bright fringe to see
the second fringe as shown by the dotted line in the figure below?
Solution:
First of all you need to realise that the central bright fringe is a result of constructive interference
and therefore the following equation is used:
dsin = m ,m = 1, 2,...
Isolate for the angle and solve:
sin = λ m/d=620e-9m*(2)/1.00e-5m=0.124
therefore, =7.12°