In a Young's double slit experiment using light of wavelength 600 nm, the two slits have unequal intensities I and 4I . Initially, the central point on the screen is a maximum. A thin transparent glass slab of refractive index 1.5 is introduced in front of one of the slits. As a result, the intensity at the central point drops to 7I . The minimum thickness of the glass slab is x 10⁻⁷ m. The value of x is _____.

Step-by-step solution

Let the phase difference introduced by the glass slab at the central point be . The new resultant intensity at the central point is given by: I_ new = I₁ + I₂ + 2 I₁ I₂ ( ) 7I = I + 4I + 2 I 4I ( ) 7I = 5I + 4I ( ) 2I = 4I ( ) ( ) = 1 2 For minimum thickness, the phase shift should be the smallest positive value, which is = 3 . The phase difference introduced by a slab of thickness t and refractive index is: = 2 ( - 1)t Substituting the known values: 3 = 2 600 10⁻⁹ (1.5 - 1)t 1 3 = 2 600 10⁻⁹ 0.5 t 1 3 = 1 600 10⁻⁹