A point source S emitting light of wavelength 600 nm is placed at a very small height h above a flat reflecting surface AB (see figure). The intensity of the reflected light is 36 % of the incident intensity. Interference fringes are observed on a screen placed parallel to the reflecting surface at a very large distance D from it. If the intensity at point P corresponds to a maximum, then the minimum distance through

Step-by-step solution

Initially, path difference at P between two waves reaching from S and S ' is 2 h Therefore, for maximum intensity at P : 2 h = n - 1 2 λ ....... (i) Now, let the sources S is moved by a distance x , the path difference will be 2 h + 2 x or 2 h - 2 x So, for displaced by x ( away or towards the mirror) then maximum intensity at P , 2 h + 2 x = n + 1 - 1 2 λ ....... (ii) 2 h - 2 x = n - 1 - 1 2 λ .......(iii) Solving Eqs. (i) and (ii) or Eqs. (i) and (iii), we get, 2 x = n+ 1 2 λ - n - 1 ⁡ 2 λ = λ x = λ 2 x = λ 2 = 6