In Young's double slit experiment, two wavelengths ( ₁=780 ~nm ) and ( ₂=520 ~nm ) are used to obtain interferance fringes. If the ( n^ th ) bright band due to ( ₁ ) concides with ( (n+1)^ th ) bright band due to ( ₂ ) then the value of ( n ) is

Step-by-step solution

Given, ( n^ th ) bright band due to ( ₁=(n+1)^ th ) bright band due to ( ₂ ) [ array l n ₁ D d = (n+1) ₂ D d n ₁=(n+1) ₂ n+1 n = ₁ ₂ array ] Given, ( ₁=780 ~nm ) and ( ₂=520 ~nm ), therefore [ n+1 n = 780 520 ] On rearranging, we get [ array l 52(n+1)=78 n 52 n+52=78 n 78 n-52 n=52 26 n=52 n= 52 26 =2 array ] Thus, value of ( n ) is ( 2 ).