Question

In biprism experiment, 10th dark band is observed at $209mm$ from the central bright point on the screen with red light of wavelength $6400\mathring{A}$. By how much will fringe width change, if blue light of wavelength $4800\mathring{A}$ is used with the same setting?

Answer 1

In the biprism experiment, the 10th dark band is observed.

The distance between the $m^{th}$ dark band with the central bright band is

$X_m=(2m-1)\frac{\lambda D}{2d}$

Therefore, the distance for the 10th dark band is

$X_{10}=\left(\right(2\times 10)-1)\frac{\lambda D}{2d}=\frac{19\lambda D}{2d}$

Now, when red light is used, we have:

$(X_{10}{)}_r=\frac{19{\lambda }_{r}D}{2d}$------(1)

Similarly, for blue light, we have

$(X_{10}{)}_b=\frac{19{\lambda }_{b}D}{2d}$------(2)

Now, the fringe width is

$X=\frac{\lambda D}{d}$

$\therefore X_r=\frac{{\lambda }_{r}D}{d}$-------(3)

$\therefore X_b=\frac{{\lambda }_{b}D}{d}$-------(4)

From equations (1) and (3), we get:

$(X_{10}{)}_r=\frac{19X_r}{2}=2.09mm$

$\therefore X_r=\frac{2\times 2.09}{19}=0.22mm$

Dividing equations (1) and (2), we get:

$\frac{(X_{10}{)}_r}{(X_{10}{)}_b}=\frac{{\lambda }_r}{{\lambda }_b}$

$\therefore (X_{10}{)}_b=1.57mm$

Now, from equations (2) and (4), we get:

$(X_{10}{)}_b=\frac{19X_b}{2}$

$\therefore X_b=\frac{2\times 1.57}{19}=0.165mm$

Therefore, the change in fringe width when blue light is used instead of red is:

$X_r-X_b=0.22-0.165=0.055mm$