Question

explain in detail position of secondary minimum

Answer 1

Consider light wave will travel by inclination ‘o’ with initial direction and converge at ‘p’. Although the incident wave on slit are in phases, but they travel on converging at ‘p’ travel unequal distances. Hence the point ‘p’ will be position of maximum & minimum intensity, which depends upon the path difference of wave converging at ‘p’. The path difference of the wave emitting from A & B and meeting at p is given by,

BN = dsinθ

The point ‘p’ will be the position of secondary minima of path difference of wave emitting from A and B and meeting at p is integral multiple of 1

 

i.e. dsinθ = nλ  where nλ= 1, 2, 3, …………………………

 

Similarly, the point ‘p’ will be the position of secondary minima of path difference of wave emitting from A and B and meeting at p is integral multiple of

i.e. dsinθ= nλ

Similarly, the point ‘p’ will be the position of secondary maxima if path difference of the wave emitting from A and B and meeting at p is odd integral multiple of 2.

i.e. dsinθ= (2n + 1)  λ/2,         n = 0, 1, 2, 3, ……………………..

Width of Central Maxima

If ‘p be the position of nth minima, then path difference,

dsinθ = nd

for nth minima since θ is very small, then Sinθ  ≈ θ

If yn be the distance of nth maxima from the centre of fringe,

The difference of consecutive distances from the center of diffractive gives the width of secondary maxima

Since, width of central maxima is double then that secondary maxima.