## 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 1M and S 2N perpendiculars from S 1 and S 2 on the screen and join S 1P and S 2P. The path difference between the waves reaching at P from S 1 and S 2 is given by

In a right angled triangle S 2NP, 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 1MP, 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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##### Topics of Interest

4.11 FRESNEL'S BIPRISM Fresnel produced the interference fringes in the laboratory by deriving two coherent sources S 1 and S 2 from a single monochromatic source S by using the deviation produced by...

4.6 ANALYTICAL TREATMENT OF INTERFERENCE We shall derive an expression for the resultant intensity at any point P of the screen due to the superposition of two waves of light. Let S be a narrow slit...

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