In Young's double-slit experiment, the intensity at a point is ( 1 4 ) of the maximum intensity. The angular position of this point is = wavelength of light used, d = separation between the two slits, (60)^ = 1 2

Step-by-step solution

The intensity at a point in Young's double-slit experiment is given by I = I₀ ^2 ( 2 ) , where I₀ is the maximum intensity and is the phase difference. Given I = I₀ 4 I₀ 4 = I₀ ^2 ( 2 ) ( 2 ) = 1 2 2 = 3 = 2 3 The phase difference is related to the path difference x by = 2 x 2 3 = 2 x x = 3 The path difference is also given by x = d d = 3 = 3d = ⁻¹ ( 3d ) Answer: ⁻¹ ( 3d )