Expression for Fringe Width in Young’s Double Slit Experiment

Class 12/Notes/Wave Optics

Introduction

In Young’s Double Slit Experiment (YDSE), two coherent light waves emerging from the slits $S_1$ and $S_2$ interfere on a screen. This produces alternate bright and dark fringes.

The distance between two successive bright fringes or two successive dark fringes is called the fringe width.

The fringe width is denoted by β (beta).

Arrangement of Young’s Double Slit Experiment

Consider two narrow slits $S_1$ and $S_2$, separated by a distance $d$. A screen is placed at a distance $D$ from the slits.

Let:

• $S_1S_2 = d$ = separation between the two slits
• $D$ = distance between the slits and the screen
• $O$ = point on the screen directly opposite the midpoint of $S_1S_2$
• $P$ = any point on the screen
• $x = OP$ = distance of point $P$ from the central point $O$
• $\theta$ = angle made by $S_1P$ with the central axis
• $\delta$ = path difference between the waves reaching $P$

For the usual YDSE arrangement,$$D \gg d$$

and the fringes are observed close to the central region of the screen.

Path Difference at Point P

The waves reaching $P$ from $S_1$ and $S_2$ travel slightly different distances.

From the geometry of the experiment, the path difference is$$\delta = S_2P-S_1P$$

For a sufficiently distant screen,$$\boxed{\delta=d\sin\theta}$$

This is the fundamental expression used to locate the interference fringes.

Small-Angle Approximation

Since the screen is far away from the double slit,$$D\gg d$$

and for points close to the central maximum, $\theta$ is small.

Therefore,$$\sin\theta\approx\tan\theta$$

From the geometry,$$\tan\theta=\frac{x}{D}$$

Hence,$$\sin\theta\approx\frac{x}{D}$$

Substituting this into the path-difference expression,$$\delta=d\frac{x}{D}$$

Therefore,$$\boxed{\delta=\frac{dx}{D}}$$

This relation connects the path difference with the position of a point on the screen.

Position of Bright Fringes

Bright fringes are produced when the two waves arrive in phase.

The condition for constructive interference is$$\delta=n\lambda$$

where$$n=0,\pm1,\pm2,\pm3,\ldots$$

Using$$\delta=\frac{dx}{D}$$

we get$$\frac{dx}{D}=n\lambda$$

Therefore,$$x_n=\frac{n\lambda D}{d}$$

Thus, the position of the $n^{\text{th}}$ bright fringe measured from the central maximum is$$\boxed{x_n=\frac{n\lambda D}{d}}$$

For the central bright fringe,$$n=0$$

and therefore, x0​=0

Position of Dark Fringes

Dark fringes are produced when the two waves arrive out of phase by π\pi.

The condition for destructive interference is$$\delta=\left(n+\frac12\right)\lambda$$

Therefore,$$\frac{dx}{D} = \left(n+\frac12\right)\lambda$$

Hence, the position of the $n^{\text{th}}$ dark fringe is xn​=(n+1/2)Dλ/D​d

Derivation of Fringe Width

The fringe width is the distance between two consecutive bright fringes or two consecutive dark fringes.

Let the positions of two successive bright fringes be$$x_n=\frac{n\lambda D}{d}$$

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

Therefore, fringe width is$$\beta=x_{n+1}-x_n$$

Substituting,$$\beta= \frac{(n+1)\lambda D}{d} – \frac{n\lambda D}{d}$$

Taking the common factor,$$\beta= \frac{\lambda D}{d} [(n+1)-n]$$

Since$$(n+1)-n=1$$

we obtain β=Dλ/​d

Expression for Fringe Width

Therefore, the expression for the fringe width in Young’s double slit experiment is$$\boxed{\displaystyle \beta=\frac{\lambda D}{d}}$$

where:

• β = fringe width
• λ = wavelength of light used
• D = distance between the double slit and screen
• d = separation between the two slits

This is one of the most important results of Young’s double slit experiment.

Fringe Width from Dark Fringes

he same result can be obtained using consecutive dark fringes.

The positions of dark fringes are$$x_n= \left(n+\frac12\right)\frac{\lambda D}{d}$$

and$$x_{n+1}= \left(n+\frac32\right)\frac{\lambda D}{d}$$

Therefore,$$\beta=x_{n+1}-x_n$$$$\beta= \left[ \left(n+\frac32\right) – \left(n+\frac12\right) \right] \frac{\lambda D}{d}$$

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

Thus, the separation between consecutive bright fringes is equal to the separation between consecutive dark fringes.

What Does the Expression Tell Us?

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

shows how the fringe width depends on the experimental parameters.

Dependence on wavelength

$$\beta\propto\lambda$$

Therefore, increasing the wavelength increases the fringe width.

A larger wavelength produces fringes farther apart.

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Dependence on screen distance

$$\beta\propto D$$

Therefore, increasing the distance between the slits and screen increases the fringe width.

Moving the screen farther away makes the fringes more widely spaced.

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Dependence on slit separation

$$\beta\propto\frac1d$$

Therefore, increasing the separation between the two slits decreases the fringe width.

Closely spaced slits produce wider fringes.

Important Result

The fringe width is independent of the order of the fringe.

For example,$$\beta_1=\beta_2=\beta_3=\cdots$$

provided the usual YDSE conditions are satisfied.

Thus, the interference pattern consists of equally spaced bright and dark fringes.

Quick Derivation Flow

The entire derivation can be remembered as:$$\boxed{\delta=d\sin\theta}$$

For small $\theta$,$$\boxed{\sin\theta\approx\tan\theta=\frac{x}{D}}$$

Therefore,$$\boxed{\delta=\frac{dx}{D}}$$

For bright fringes,$$\delta=n\lambda$$

so,$$\boxed{x_n=\frac{n\lambda D}{d}}$$

Hence,$$\beta=x_{n+1}-x_n$$

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

Key Formulae at a Glance

Quantity	Expression

Path difference

$\delta=d\sin\theta$

Small-angle approximation

$\sin\theta\approx x/D$

Path difference near central region

$\delta=dx/D$

Position of $n^{th}$ bright fringe

$x_n=n\lambda D/d$

Position of $n^{th}$ dark fringe

$x_n=(n+\frac12)\lambda D/d$

Fringe width

β=λD/d​

Conditions for the Standard Fringe-Width Expression

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

is obtained under the usual conditions of Young’s double slit experiment:

1. The two slits act as coherent sources.
2. The wavelength $\lambda$ remains constant.
3. The slit separation $d$ is much smaller than the screen distance $D$.
4. The observation is made close enough to the central region for the small-angle approximation to be valid.
5. The two interfering waves have comparable amplitudes for clearly visible fringes.

Worked Example

Example

In a Young’s double slit experiment, monochromatic light of wavelength $600\,\text{nm}$ is used. The distance between the slits is $0.5\,\text{mm}$, and the screen is $2\,\text{m}$ away. Find the fringe width.

Given

$$\lambda=600\times10^{-9}\,\text{m}$$$$D=2\,\text{m}$$$$d=0.5\times10^{-3}\,\text{m}$$

Using$$\beta=\frac{\lambda D}{d}$$$$\beta= \frac{600\times10^{-9}\times2} {0.5\times10^{-3}}$$$$\boxed{\beta=2.4\times10^{-3}\,\text{m}}$$

Therefore,$$\boxed{\beta=2.4\,\text{mm}}$$

So, consecutive bright fringes—or consecutive dark fringes—are separated by 2.4 mm.

Exam Tip

A very common question in NEET and JEE is to ask how the fringe width changes when $\lambda$, $D$, or $d$ is changed.

Remember the relationship:$$\boxed{\beta\propto\frac{\lambda D}{d}}$$

Therefore:

• $\lambda$ increases → β increases
• $D$ increases → β increases
• $d$ increases → β decreases

One-line memory trick

“Fringe width follows wavelength and screen distance, but opposes slit separation.”

Final Result

For Young’s double slit experiment, the path difference at a point $P$ on the screen is$$\delta=d\sin\theta$$

and, for small angles,$$\delta=\frac{dx}{D}$$

The positions of successive bright fringes are$$x_n=\frac{n\lambda D}{d}$$

Therefore, the distance between two successive bright fringes is$$\boxed{\displaystyle \beta=\frac{\lambda D}{d}}$$

The same expression is obtained for successive dark fringes.

Fringe width in Young’s double slit experiment:$$\boxed{\beta=\frac{\lambda D}{d}}$$

The fringe width increases with wavelength and screen distance, and decreases with slit separation.