Question

  In one of Young's double slit experiment, two thin parallel slits S1 and Sz, distant a apart are

          illuminated with by an intermediate slit S parallel to S1 and S2 and situated at the same distance

          from each. A system of fringes is produced on a screen P, parallel to the double slit and distant D

           from it. It is discovered that N consecutive fringes of the same nature occupy a length L.

           (i.) Derive, from first principle, the expression for λ, the wavelength of the light used, in terms of

           a, D, l and N. Hence, calculate λ given that a = 2mm, D = 1m, l = 4mm, N = 12.

           (ii.) What will be the new value of L, for the same number N of fringes if the apparatus is plunged in a pool of water? (Hint: index of water= 4/3)

          

Answer 1

As we know in the double slit experiment ,(take a=d here)

when d<<D

$\Delta x=dsin\theta$ ...................1

and when observation is near to the center or say angle is very less than,

$\Delta x= {dy\over D}$ ...............................2

and from 1 and 2 we derived a frendlwich width$(\beta)$

$\boxed{\beta={\lambda D\over d}}$

so as given N as consecutive finges ,here we take these as maximas,

so formula for length L will be,

$\boxed{L=N\beta\\}\\\boxed{L={N\lambda D\over d}}$

therefore ,$\boxed{\lambda={Ld \over ND}}$

a)now putting given values in above formula we get,

$\lambda=6.667*10^{-7}$

b)there will be n change in the value L as

,$\boxed{L={\lambda ND \over d}}$

and here,$\lambda \implies \mu \lambda$

therefore L will become $4\over3$ L.