In Young's double slit experiment, slits are separated by ( 2 ~mm ) and the screen is placed at a distance of ( 1.2 ~m ) from the slits. Light consisting of two wavelengths ( 6500 Å ) and ( 5200 Å ) are used to obtain interference fringes. Then the separation between the fourth bright fringes of two different patterns produced by the two wavelengths is

Step-by-step solution

We know that, y= n D d where is wavelength; D is distance of screen from slits; d is separation between the slits. So, y₁= n ₁ D d and y₂= n ₂ D d Therefore, y₁-y₂= n D d ( ₁- ₂ ) Given, n=4 ; D=1.2 ~m ; d=2 ~mm =2 10⁻³ ~m ; ₁=6500 Å=6500 10⁻¹⁰ ~m ; ₂=5200 Å=5200 10⁻¹⁰ ~m y₁-y₂= 4 1.2 2 10⁻³ (6500 10⁻¹⁰-5200 10⁻¹⁰ )=2.4 10³ 1300 10⁻¹⁰ ~m =3120 10⁻⁷=0.3120 10⁻³ ~m y₁-y₂=0.3120 ~mm Therefore, the separation between the fourth bright fringesof two different patterns produced by the two wavelengths is 0.3120 ~mm .