Derive an expression for path difference in Young’s double slit experiment and obtain the conditions for constructive and destructive interference at a point on the screen.

#### Solution

Diagrammatical arrangement for Young’s double slit experiment:

In the above figure, S_{1} and S_{2} are two narrow closely spaced slits illuminated by monochromatic light of wavelength λ. The screen on which the interference pattern is observed is represented as XY.

If S_{1} and S_{2} emit light in the same phase, then for point O, the path difference receives light in the same phase. The superposition at O is constructive producing a bright point called the central maxima. The intensity at any point P at a distance x from O depends on the path difference between light reaching P from S_{1} and S_{2}. The path difference is S_{2}P − S_{1}P.

From the geometry of the figure, we have

`(S_2P)^2=D^2+(x+d/2)^2`

Similarly, we have

`(S_1P)^2=D^2+(x+d/2)^2`

`:.(S_2P)^2-(S_1P)^2=D^2+(x+d/2)^2-D^2-(x-d/2)^2`

`:.(S_2P)^2-(S_1P)^2=(x+d/2)^2-(x-d/2)^2`

`:.(S_2P)^2-(S_1P)^2=x^2+xd+d^2/4-x^2+xd-d^2/4`

∴ (S_{2}P)^{2}-(S_{1}P)^{2}=2xd

∴(S_{2}P-S_{1}P)(S_{2}P+S_{1}P)=2xd

`:.S_2P-S_1P=(2xd)/(S_2P+S_1P)`

Now, from the figure we can see that

S_{2}P≈S_{1}P = D

`:.S_2P-S_1P=(2xd)/(2D)=(xd)/D`

This is the expression for path difference.

**Condition for constructive interference**: Constructive interference will occur when the phase difference between the two superposing waves is an even multiple of π or the path difference is an integral multiple of wavelength λ.

**Condition for destructive interference**: Destructive interference will occur when the phase difference between the two superposing waves is an odd multiple of π or the path difference is an odd multiple of wavelength λ/2.