Q.In a biprism experiment, the 10th dark band is observed at 2.09 mm from the central bright point on the screen with red light of wavelength 6400 Å. By how much will the fringe width change if blue light of wavelength 4800 Å is used with the same setting?

In a biprism experiment, the $n$-th dark fringe from the centre is at distance $(2n-1)\beta/2$; use this to first find $\beta$ for red light, then rescale it for blue light since $\beta\propto\lambda$.

In a biprism (or any two-slit-type) interference pattern, dark fringes occur at odd multiples of half the fringe width from the central bright fringe. The position of the $n$-th dark band from the centre is

$x_n=(2n-1)\frac{\beta}{2}$

For the 10th dark band ($n=10$) at $x_{10}=2.09$ mm:

$2.09=(2\times10-1)\frac{\beta_1}{2}=19\times\frac{\beta_1}{2}=9.5\,\beta_1$

$\beta_1=\frac{2.09}{9.5}=0.22\text{ mm}$

This is the fringe width for red light ($\lambda_1=6400$ Å).

Since the fringe width $\beta=\dfrac{\lambda D}{d}$ is directly proportional to the wavelength (same $D$, $d$ for the unchanged setup): …