In a Young's double slit experiment using light of wavelength 500 ~nm , a thin transparent glass slab of refractive index 1.5 is placed in front of one of the slits. As a result, the intensity of light at the geometric center of the screen becomes exactly half of the maximum intensity. The minimum possible thickness of the glass slab is ________ nm .

Step-by-step solution

The intensity I at any point is given by: I = I_ ^2 ( 2 ) Given that the intensity at the geometric center is I_ 2 : I_ 2 = I_ ^2 ( 2 ) ^2 ( 2 ) = 1 2 For the minimum thickness, we consider the smallest phase difference: 2 = 4 = 2 The corresponding path difference x is: x = 2 = 2 ( 2 ) = 4 Given = 500 ~nm : x = 500 4 = 125 ~nm The optical path difference introduced by a glass slab of thickness t and refractive index is: x = ( - 1)t Equating the two expressions for path difference: ( - 1)t = 125 Given = 1.5 : (1.5 -