91Ó°ÊÓ

Wavelength emitted: $502\mathrm{nm}$

Distance between slit and screen:$120\mathrm{m}$

Distance of 20th fringe from the center of central bright fringe: $10.6\mathrm{mm}$

The double Slit experiment is a simple but substantial demonstration that suggests that both light and matter can exhibit characteristics of waves and particles under the constraints of classical definition.

The expression connecting all the parameters of the experiment is,

${\mathbf{y}}_{\mathbf{m}}\mathbf{=}\mathbf{R}\frac{\mathbf{λ³¾}}{\mathbf{d}}$ ...(i)

Here, ${\mathbf{y}}_{\mathbf{m}}$is the distance of the fringe, $\mathbf{λ}$is the wavelength, m is the order of the fringe,

R is the distance between the screen and the slits and d is the slit width.

The distance of the 20th fringe is given as,

${\mathrm{y}}_{20}=10.6\mathrm{x}{10}^{-3}\mathrm{m}$

From equation (i), solve for the separation between slits,

$$\begin{gathered} \mathrm{d}=\frac{20\mathrm{¸éλ}}{{\mathrm{y}}_{20}} \\ =\frac{\left(20\right)\left(1.20\mathrm{m}\right)\left(502\mathrm{x}{10}^{-9}\mathrm{m}\right)}{10.6\mathrm{x}{10}^{-3}\mathrm{m}} \\ =1.14\mathrm{x}{10}^{-3}\mathrm{m} \\ =1.14\mathrm{mm} \end{gathered}$$

Thus, the distance between two slits is 1.14 m