Question

In a double-slit experiment, the slit separation is 2.0 mmmm , two wavelengths of 910 nmnm and 650 nmnm illuminate the slits, the screen is placed 2.8 meters away from the slits. At what distance from the central maximum on the screen will a dark fringe from one pattern first coincide with a dark fringe from the other

Answer 1

Answer:

Approximately $3.2\; {\rm mm}$.

Explanation:

In diffraction, dark fringes (minimum) are observed when light from the two slits interfere destructively. These interferences require a phase difference of $(1/2)$ cycle. At the $n$th minimum, the path difference would needs to be $(n + (1/2))\, \lambda$ (for integer values of $n$.)

Let $\lambda_{a} = 910\; {\rm nm}$ and $\lambda_{b} = 650\; {\rm nm}$ denote the wavelength of the two waves. Assume that the $a$th minimum of the $\lambda_{a} = 910\; {\rm nm}\!$ wave coincides with the $b$th minimum of the $\lambda_{b} = 650\; {\rm nm}\!$ wave. (Both $a$ and $b$ are non-negative integers.)

The path difference of the two waves need to match:

$\displaystyle \left(a + \frac{1}{2}\right)\, \lambda_{a} = \left(b + \frac{1}{2}\right)\, \lambda_{b}$.

$\displaystyle \left(a + \frac{1}{2}\right)\, 910\; {\rm nm} = \left(b + \frac{1}{2}\right)\, 650\; {\rm nm}$.

In other words, the value of non-negative integers $a$ and $b$ need to satisfy:

$910\, a + 455 = 650\, b + 325$.

There might be more than one pairs of $a$ and $b$ that satisfy the constraints. In general, the least positive $a\!$ that meets the requirements can be found using linear programming techniques.

Specifically in this example, note that:

$910\, a + 455 = 650\, b + 325$.

$70\, a+ 35 = 50\, b + 25$.

$14\, a + 7 = 10\, b + 5$.

The least positive value of $a$ that satisfy the requirements is $a = 2$, for which $b = 3$.

At the position where minimum of the two waves coincide, path difference would be:

• $\displaystyle (2 + (1/2))\, (910\times 10^{-9}\; {\rm m}) = 2.275 \times 10^{-6}\; {\rm m}$ for the $\lambda_{a} = 910\; {\rm nm}$ wave, which is the same as
• $\displaystyle (3 + (1/2))\, (650\times 10^{-9}\; {\rm m}) = 2.275 \times 10^{-6}\; {\rm m}$ for the $\lambda_{a} = 650\; {\rm nm}$ wave.

In double-slit diffraction, for a pattern generated from a path difference of $m\, \lambda$, the angle $\theta$ between that pattern and the central maximum (relative to the center of the slits) should satisfy:

$\displaystyle \sin(\theta) = \frac{m\, \lambda}{d}$,

Where $d$ is the distance between (the center of) the two slits.

Rearrange to obtain:

$\displaystyle \theta = \arcsin\left(\frac{m\, \lambda}{d}\right)$.

In this question, the path difference is $2.275 \times 10^{-6}\; {\rm m}$, while the distance between the two slits is $2.0 \; {\rm mm} = 2.0 \times 10^{-3}\; {\rm m}$. The angle between the pattern on the screen and the central maximum should be:

$\begin{aligned} \theta &= \arcsin\left(\frac{m\, \lambda}{d}\right) \\ &= \arcsin\left(\frac{2.275 \times 10^{-6}\; {\rm m}}{2.0 \times 10^{-3}\; {\rm m}}\right) \\ &\approx 0.06517^{\circ}\end{aligned}$.

Since the screen is at a distance of $L = 2.8\; {\rm m}$ from the screen, the on-screen distance $Z$ between the central maximum and the pattern at $\theta \approx 0.06517^{\circ}$ would be:

$\begin{aligned} Z &= L\, \tan(\theta) \\ &\approx (2.8\; {\rm m})\, \tan(0.06517^{\circ}) \\ &\approx 3.2 \times 10^{-3}\; {\rm m}\\ &\approx 3.2\; {\rm mm} \end{aligned}$.