Question

Obtain an expression for fringe width in Young's double-slit experiment.

Hint:

To increase the fringe width (make the pattern more spread out), you can either increase the distance to the screen ($D$), use light with a longer wavelength ($\lambda$), or bring the two slits closer together (decrease $d$).

Answer 1

Step 1: Experimental Setup.
Consider two coherent sources $S_1$ and $S_2$ separated by a distance $d$. Let $D$ be the distance between the slits and the screen. A point $P$ is located on the screen at a distance $y$ from the central maximum $O$.

Step 2: Finding the Path Difference.
The path difference between the waves reaching $P$ from $S_1$ and $S_2$ is $\Delta p = S_2P - S_1P$. For $D \gg d$, this path difference is approximately: $$\Delta p = \frac{dy}{D}$$ For a bright fringe (constructive interference), the path difference must be an integral multiple of the wavelength ($\lambda$): $$n\lambda = \frac{dy_n}{D} \implies y_n = \frac{nD\lambda}{d}$$ where $y_n$ is the position of the $n^{th}$ bright fringe.

Step 3: Calculating Fringe Width.
The fringe width ($\beta$) is the distance between two consecutive bright or dark fringes. It is calculated by subtracting the position of the $n^{th}$ fringe from the $(n+1)^{th}$ fringe: $$\beta = y_{n+1} - y_n$$ $$\beta = \frac{(n+1)D\lambda}{d} - \frac{nD\lambda}{d}$$ $$\beta = \frac{D\lambda}{d}$$ This expression shows that the fringe width is constant for a given experimental setup.