Young’s Double Slit Experiment

YOUNG'S DOUBLE SLIT EXPERIMENT

A train of plane light waves is incident on a barrier containing two narrow slits separated by a distance 'd'. The widths of the slits are small compared with wavelength of the light used, so that interference occurs in the region where the light from S₁ overlaps that from S₂. A series of alternately bright and dark bands can be observed on a screen placed in this region of overlap.

The variation in light intensity along the screen near the centre O shown in the figure

Now consider a point P on the screen. The phase difference between the waves at P is $\varphi$, where

$\varphi$= $\frac{2\pi }{\lambda }\Delta P_o$

(where $\Delta$P_o is optical path difference, $\Delta$P_o=$\mu$$\Delta$P_g; $\Delta$P_g being the geometrical path difference.)

$=\frac{2\pi }{\lambda }\left[S_{2}P-S_{1}P\right]$ (here $\mu$ = 1 in air) As As, D >> d,

S₂P - S₁P $\approx$ d sinq

sin $\theta$ $\approx$ tan$\theta$( = y/D).

[for very small $\theta$]

Thus, $\varphi$ = $\frac{2\pi }{\lambda }\left(\frac{dy}{D}\right)$

For constructive interference,

$\varphi$ = 2n$\pi$ (n = 0, 1, 2...)

$\Rightarrow$ $\frac{2\pi }{\lambda }\left(\frac{dy}{D}\right)=2n\pi$ $\Rightarrow$ y = n$\frac{\lambda D}{d}$

Similarly for destructive interference,

y = (2n - 1)$\frac{\lambda D}{2d}$ (n = 1, 2, .........)

Fringe Width W

It is the separation of two consecutive maxima or two consecutive minima.

Near the centre O [where $\theta$ is very small],

W = y_n+1 - y_n [y_n gives the position of nth maxima on screen]

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

Intensity Variation on Screen.

If A and I_o represent amplitude of each wave and the associated intensity on screen, then, the resultant intensity at a point on the screen corresponding to the angular position q as in above figure, is given by

I = I_o­ + I_o + 2$\sqrt{I_o^2}$ cos$\varphi$, when $\varphi$ = $\frac{2\pi \left(d\sin \theta \right)}{\lambda }$= 4I_o cos²$\frac{\varphi }{2}$

Illustration 1: A beam of light consisting of two wavelengths 6500 ^oA and 5200 ^oA is used to obtain interference fringes in YDE. The distance between the slits is 2.0 mm and the distance between the plane of the slits and the screen is 120 cm.

(a) Find the distance of the third bright fringe on the screen from the central maxima for the wavelength 6500 ^oA.

(b) What is the least distance from the central maxima where the bright fringes due to both the wavelengths coincide?

Solution: (i) $y_3=n.\frac{D\lambda }{d}=\frac{3x1.2mx6500x{10}^{-10}m}{2x{10}^{-3}m}=0.12cm$

Let nth maxima of light with wavelength 6500 Å coincides with that of mth maxima of 5200Å.

(ii) $\frac{mx6500A^{o}xD}{d}=\frac{nx5200A^{o}xD}{d}\Rightarrow \frac{m}{n}=\frac{5200}{6500}=\frac{4}{5}$

$\text{Least distance = }{\text{y}}_{\text{4}}\text{=}\frac{\text{4}\text{.D}\text{(6500 }{\text{A}}^{\text{o}}\text{)}}{\text{d}}$ =$\frac{4x6500x{10}^{-10}x1.2}{2x{10}^{-3}m}=0.16cm$

Illustration 2: The intensity of the light coming from one of the slits in a Young's double slit experiment is double the intensity from the other slit. Find the ratio of the maximum intensity to the minimum intensity in the interference fringe pattern observed.

Solution : $\frac{I_{\max }}{I_{\min }}={\left[\frac{\sqrt{l_1}+\sqrt{l_2}}{\sqrt{l_1}-\sqrt{l_2}}\right]}^{2}As{I}_1=2I_2\Rightarrow \frac{I_{\max }}{I_{\min }}={\left[\frac{\sqrt{2}+1}{\sqrt{2}-1}\right]}^2=34$

Displacement of Fringes

When a film of thickness 't' and refractive index '$\mu$' is introduced in the path of one of the sources, then fringe shift occurs as the optical path difference changes.

Optical path difference at

P = S₂P - [S₁P+ $\mu$t - t] = S₂P - S₁P - ($\mu$ - 1)t = y.d/D - ($\mu$ - 1)

$\Rightarrow \text{nth fringe is shifted by }\Delta y=\frac{D(\mu -1)t}{d}=\frac{w}{\lambda }(\mu -1)t$