Young’s Double Slit Experiment

 

Young’s Double Slit Experiment, Types of fringes

According to Huygens’ principle, every point on a wavefront acts as a source of secondary wavelets. In Young’s experiment, light from a monochromatic source illuminates two narrow slits $S_1$ and $S_2$. 

These two slits behave as coherent sources because the waves emerging from them have:

• The same frequency
• A constant phase difference
• The same wavelength

When the waves from $S_1$ and $S_2$ overlap on the screen, they produce interference fringes consisting of alternate bright and dark bands.

When two waves travel through the same medium simultaneously, the resultant displacement at any point is equal to the algebraic sum of the individual displacements produced by each wave.

If the displacements due to two waves are:

$$y_1=a_1\sin \omega t$$

and

$$y_2=a_2\sin(\omega t+\phi)$$

then, according to the principle of superposition:

$$\boxed{y=y_1+y_2}$$

where $\mathrm{y\; is\; the\; resultant\; displacement.}$

For two coherent waves of equal amplitude $\mathrm{a:}$

$$y_1=a\sin \omega t$$$$y_2=a\sin(\omega t+\phi)$$

Using the trigonometric identity:

$$\sin A+\sin B =2\sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$$

we get:

$$y=2a\cos\frac{\phi}{2} \sin\left(\omega t+\frac{\phi}{2}\right)$$

Therefore, the amplitude of the resultant wave is:

$$\boxed{A=2a\cos\frac{\phi}{2}}$$

The intensity is proportional to the square of amplitude:

$$I\propto A^2$$

Therefore:

$$\boxed{I=4I_0\cos^2\frac{\phi}{2}}$$

where $I_0$ is the intensity due to either wave individually.

Case I: Constructive Interference:

$$\phi=2n\pi$$
$$\Delta=n\lambda$$

The resultant amplitude is maximum:

$$A=2a$$

and the fringe appears bright.

Case II: Destructive Interference:

$$\phi=(2n+1)\pi$$
$$\Delta=\frac{(2n+1)\lambda}{2}$$

The resultant amplitude is minimum:

$$A=0$$

and the fringe appears dark.

Fringe Width

Consider two coherent slits $S_1$ and $S_2$, separated by distance $d$. Let the screen be at a distance $D$ from the slits.

Let $\mathrm{P\; be\; a\; point\; on\; the\; screen\; at\; a\; distance }$$y$ from the central point $\mathrm{O.}$

The path difference between the waves reaching $\mathrm{P\; is:}$

$$\Delta = S_2P-S_1P$$

From the geometry of the experiment:

$$\Delta=d\sin\theta$$

For small angles:

$$\sin\theta \approx \tan\theta$$

Therefore,

$$\sin\theta \approx \frac{y}{D}$$

Hence,

$$\boxed{\Delta=\frac{dy}{D}}$$

(i) Bright Fringes

For constructive interference:

$$\Delta=n\lambda$$

where $n=0,1,2,3,\dots$

Therefore,

$$\frac{dy_n}{D}=n\lambda$$

Thus,

$$\boxed{y_n=\frac{n\lambda D}{d}}$$

This gives the position of the $n^{\text{th}}$ bright fringe.

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(i) Dark Fringes

For destructive interference:

$$\mathrm{\Delta }=\frac{(2n+1)\lambda }{\mathrm{2}}$$

Therefore,

$$\frac{dy_n'}{D}=\frac{(2n+1)\lambda}{2}$$

Hence,

$$\boxed{y_n'=\frac{(2n+1)\lambda D}{2d}}$$

This gives the position of the $n^{\text{th}}$ dark fringe.

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Derivation of Fringe Width

The position of two successive bright fringes is:

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

and

$$y_{n+1}=\frac{(n+1)\lambda D}{d}$$

Fringe width is:

$$\beta=y_{n+1}-y_n$$
$$\beta=\frac{(n+1)\lambda D}{d} -\frac{n\lambda D}{d}$$ $$\boxed{\beta=\frac{\lambda D}{d}}$$

Therefore, the fringe width in Young’s double-slit experiment is:

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

Where:

• $\beta$ = fringe width
• $\lambda$ = wavelength of light
• $\mathrm{D\; =\; distance\; between\; slits\; and\; screen}$
• $d$ = separation between the two slits