Question

Coherent light with wavelength = 600 nm falls on two very narrow closely spaced slits and the interference pattern is observed on a screen that is 4 m from the slits. Near the center of the secreen the separation between adjacent maxima is 2 mm. What is the distance between the two slits?

Answer 1

Answer:

The distance between the two slits is 1.2mm.    

Explanation:

The physicist Thomas Young establishes, through its double slit experiment, a relationship between the interference (constructive or destructive) of a wave, the separation between the slits, the distance between the two slits to the screen and the wavelength.

$\Lambda x = L\frac{\lambda}{d}$  (1)

Where $\Lambda x$ is the distance between two adjacent maxima, L is the distance of the screen from the slits, $\lambda$ is the wavelength and d is the separation between the slits.  

If light pass through two slits a diffraction pattern in a screen will be gotten, at which each bright region corresponds to a crest, a dark region to a trough, as consequence of constructive interference and destructive interference in different points of its propagation to the screen.  

Therefore, d can be isolated from equation 1.

$d = L\frac{\lambda}{\Lambda x}$  (2)

Notice that it is necessary to express L and $\lambda$ in units of millimeters.

$L = 4m \cdot \frac{1000mm}{1m}$ ⇒ $4000mm$

$\lambda = 600nm \cdot \frac{1mm}{1x10^{6}nm}$ ⇒ $0.0006mm$

$d = (4000mm)\frac{0.0006mm}{2mm}$

$d = 1.2mm$

Hence, the distance between the two slits is 1.2mm.