Question

How do you calculate fringe spacing?

Answer 1

Fringe spacing:

1. The distance between two successive bright or dark fringes is known as the fringe spacing or fringe width. All of the fringes in the Youngs Double Slit experiment are the same length.
2. The formula for fringe spacing:

$\Rightarrow ∆\mathrm{x}=\frac{\mathrm{\lambda L}}{\mathrm{d}}$

Where,

$$\begin{gathered} \Rightarrow ∆\mathrm{x}=\mathrm{fringe}\mathrm{width} \\ \Rightarrow \lambda =\mathrm{wavelength}\mathrm{of}\mathrm{light} \\ \Rightarrow \mathrm{L}=\mathrm{Distance}\mathrm{between}\mathrm{the}\mathrm{source}\mathrm{and}\mathrm{screen} \\ \Rightarrow \mathrm{d}=\mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{two}\mathrm{sources} \end{gathered}$$

Young's Double Slit Experiment:

1. Thomas Young produced the first actual demonstration of optical interference in 1801.
2. His research provided highly convincing evidence in favor of the light wave theory.

Experimental Arrangement:

1. $S$ is a slit that gets illumination from a monochromatic light source. Since slit $S$ is small, it diffracts light, causing it to fall on slits $A$ and $B$.
2. Two waves interfere with one another on the screen after traveling through the two apertures. A and B's slits provide a consistent source of light. On the screen, bright and black fringes alternate due to wave interference.

Step 1: Given that

Let,

$$\begin{gathered} \mathrm{\lambda }=\mathrm{wavelength}\mathrm{of}\mathrm{light} \\ \mathrm{d}=\mathrm{distance}\mathrm{between}\mathrm{slits}\mathrm{A}\mathrm{and}\mathrm{B} \\ \mathrm{L}=\mathrm{distance}\mathrm{between}\mathrm{slits}\mathrm{and}\mathrm{screen} \end{gathered}$$

Step 2: Calculate the path difference,

Consider a point $P$ on the screen where light waves from slits $A$ and $B$ interfere, causing $PC=y.$ The light wave from $A$ travels a distance of$AP=r_1$, and the light wave from $B$ travels a distance of $BP=r_2$, causing $PB$ to be higher than $AP$.

$$\begin{gathered} \Rightarrow \mathrm{Path}\mathrm{difference}=\mathrm{BP}-\mathrm{AP}\Rightarrow \mathrm{BD} \\ \Rightarrow \mathrm{S}={\mathrm{r}}_2-{\mathrm{r}}_1 \end{gathered}$$

Therefore,

$\begin{array}{lcl}& & \mathrm{In}\mathrm{right}\mathrm{angle}∆\mathrm{BAD}\\ & \Rightarrow & \sin \left(\mathrm{\theta }\right)=\frac{\mathrm{BD}}{\mathrm{AB}}\\ & \Rightarrow & \sin \left(\mathrm{\theta }\right)=\frac{\mathrm{s}}{\mathrm{d}}\\ & \Rightarrow & \mathrm{s}=\mathrm{dsin}\left(\mathrm{\theta }\right)........\left(1\right)\end{array}$

Since the value of $d$ is very small as compared to $L$, therefore, $\left(\theta \right)$ will also be very small. In this condition, we can assume that :

$$\begin{gathered} \Rightarrow \sin \left(\mathrm{\theta }\right)=\tan \left(\mathrm{\theta }\right)\left(\mathrm{from}\mathrm{equation}1\right) \\ \Rightarrow \mathrm{s}=\mathrm{dtan}\left(\mathrm{\theta }\right)........\left(2\right) \\ \therefore \mathrm{In}\mathrm{right}\mathrm{angle}∆\mathrm{PEC} \\ \Rightarrow \tan \left(\mathrm{\theta }\right)=\frac{\mathrm{PC}}{\mathrm{Ec}}\Rightarrow \frac{\mathrm{y}}{\mathrm{L}} \end{gathered}$$

Step 3: Further solve the above expression,

Put the value of $\tan \left(\mathrm{\theta }\right)\mathrm{in}\mathrm{equation}\left(2\right)$, we get

$$\begin{gathered} \Rightarrow \mathrm{s}=\frac{\mathrm{dy}}{\mathrm{L}} \\ \Rightarrow \mathrm{y}=\frac{\mathrm{sL}}{\mathrm{d}}.......\left(3\right) \end{gathered}$$

Step 4: Calculation for bright fringe,

Put the value in $equation\left(3\right)$

$$\begin{gathered} \Rightarrow \mathrm{s}=\mathrm{m\lambda }\left[\mathrm{put}\mathrm{in}\mathrm{equation}\left(3\right)\right] \\ \Rightarrow \mathrm{y}=\frac{\mathrm{m\lambda L}}{\mathrm{d}}[where"m"isaninteger] \end{gathered}$$

Step 5: Calculation for dark fringe at P,

$$\begin{gathered} \Rightarrow \mathrm{s}=\left(\mathrm{m}+\frac{1}{2}\right)\mathrm{\lambda }\left[\mathrm{put}\mathrm{in}\mathrm{equation}\left(3\right)\right] \\ \Rightarrow \mathrm{y}=\frac{\left(\mathrm{m}+\frac{1}{2}\right)\mathrm{\lambda L}}{\mathrm{d}} \end{gathered}$$

Step 6: Calculate fringe spacing:

Fringe spacing is the separation between any two consecutive bright fringes and any two consecutive dark fringes. Equal amounts of dark or brilliant fringes have the same spacing or thickness. It's indicated by $∆x$.

Consider bright fringe,

$$\begin{gathered} \Rightarrow \mathrm{y}=\frac{\mathrm{m\lambda L}}{\mathrm{d}} \\ \mathrm{For}\mathrm{bright}\mathrm{fringe},\mathrm{m}=1 \\ \Rightarrow {\mathrm{y}}_1=\frac{\left(1\right)\mathrm{\lambda L}}{\mathrm{d}} \\ \Rightarrow \mathrm{for}\mathrm{next}\mathrm{order}\mathrm{bright}\mathrm{fringe},\mathrm{m}=2 \\ \Rightarrow {\mathrm{y}}_2=\frac{\left(2\right)\mathrm{\lambda L}}{\mathrm{d}} \end{gathered}$$

Step 7: Further solve the above expression for fringe spacing,

$$\begin{gathered} \Rightarrow \mathrm{fringe}\mathrm{spacing}={\mathrm{y}}_2-{\mathrm{y}}_1 \\ \Rightarrow ∆\mathrm{x}=\frac{\left(2\right)\mathrm{\lambda L}}{\mathrm{d}}-\frac{\left(1\right)\mathrm{\lambda L}}{\mathrm{d}} \\ \Rightarrow ∆\mathrm{x}=\frac{\left(2-1\right)\mathrm{\lambda L}}{\mathrm{d}} \\ \Rightarrow ∆\mathrm{x}=\frac{\mathrm{\lambda L}}{\mathrm{d}} \end{gathered}$$