Question

Young’s experiment is performed with light from excited helium atoms ($\mathbf{\lambda }$= 502 nm). Fringes are measured carefully on a screen 1.20 m away from the double slit, and the center of the 20th fringe (not counting the central bright fringe) is found to be 10.6 mm from the center of the central bright fringe. What is the separation of the two slits?

Short answer

The separation at which the two slits are situated is 1.14 mm.

Step-by-step solution

Given Data

Wavelength emitted: $502\mathrm{nm}$

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

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

(a) Concept of Young’s double-slit experiment.

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,

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

Here, ${\mathbf{y}}_{\mathbf{m}}$is the distance of the fringe, $\mathbf{\lambda }$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.

(b) Determination of the separation between the two slits.

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,

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

Thus, the distance between two slits is 1.14 m