Discussion Forum

In the biprism experiment, the distance between source and eyepiece is 1.2 m, the distance between two virtual sources is 0.84 mm. Then the wavelength of light  used if the eyepiece is to be moved transversely through a distance of 2.799 cm to shift 30 fringes is 

Answer:

Explanation:

 In a biprism  experiment, wavelength of the light is given as 

                $\lambda=\frac{d\beta}{D}$  .......(i)

 where, $\beta$ is the fringe width, 

d is the distance between the two sources and D is the distance between the source and eyepeice.

 Given  , D=1.2 m

    d= 0.84 mm=0.84 x 10^-3 m and 

   $\beta=\frac{2.799\times 10^{-2}}{30}=9.33 \times 10^{-4}$

  Substituting these values in Eq.(i) , we get

$\lambda=\frac{0.84 \times 10^{-3}\times9.33 \times 10^{-4}}{1.2}=6.531 \times 10^{-7}m =$

     =6531  Å