##### From Engineering Physics: Fundamentals and Modern Applications

## 4.10 DIVISION OF WAVEFRONT

### Theory of Interference Fringes

Let S be a narrow slit illuminated by monochromatic light of wavelength ?, and S _{1} and S _{2} be the two narrow slits separated by a small distance 2 *d* (as in Figure 5). Let a screen XY be placed at a distance D from the coherent sources S _{1} and S _{2}. Let O be the foot of the perpendicular drawn from C, the mid point of S _{1} and S _{2} on the screen. Now consider a point P on the screen at a distance *x* from O at which the condition for bright and dark fringes are to be determined. Draw S _{1}M and S _{2}N perpendiculars from S _{1} and S _{2} on the screen and join S _{1}P and S _{2}P. The path difference between the waves reaching at P from S _{1} and S _{2} is given by

In a right angled triangle S _{2}NP, we have

Figure 5

Using the binomial theorem and neglecting higher terms, since D >> *x* or *d*, we get

(1) | |

Similarly in a right angled triangle S _{1}MP, we have

(2) | |

? Path difference,

(3) | |

If the sources S _{1} and S _{2} are in the same phase as the waves reaching at P, the path differences must only be due to path difference, ?

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