Class 11 > Unit # 13: Physical Optics > Young's Double Slit Experiment

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Young’s Double Slit Experiment & Fringe Spacing Derivation

Video Lecture: Young’s Double Slit Experiment

Watch the complete step-by-step video lecture explaining Young's Double Slit Experiment and the derivation for fringe spacing:

Young’s Double Slit Experiment

“The fringes obtained by Young’s double slit experiment are the result of interference of light waves from two coherent sources.”

Essential Conditions:

1. Monochromatic light must be used so that all waves share a single, uniform wavelength.
2. The slits must be narrow (on the order of the wavelength of light) and act as coherent sources.
3. The two sources must be very close to each other because the wavelength of light is extremely small; otherwise, the bright and dark patterns would be too fine to resolve and observe clearly.

Light from a monochromatic source illuminates slit $S$ and subsequently passes through two narrow, parallel slits $S_1$ and $S_2$ separated by a small distance $d$. Since light diverging from $S_1$ and $S_2$ originates from a common wavefront, they act as two close coherent sources. An interference pattern consisting of alternating bright and dark fringes is formed on a screen placed at a distance $L$ from the double slits.

If $O$ is the central point on the screen, the path difference between $S_1O$ and $S_2O$ is zero, resulting in constructive interference at $O$ (the central bright fringe).

Fig: Experimental setup of Young's double slit experiment showing light rays traveling to point $P$ on the screen.

Path Difference at Any Point P

Consider any point $P$ on the screen at a distance $Y$ from the central maximum $O$. Waves traveling from $S_2$ to $P$ cover a distance $r_2 = S_2P$, while waves from $S_1$ cover a distance $r_1 = S_1P$. The intensity of light at $P$ is determined by the superposition of these waves.

The path difference between the waves from $S_1$ and $S_2$ is given by:

$\text{Path Difference} = r_2 - r_1 = d \sin\theta$

Conditions for Maxima and Minima

1. Constructive Interference (Maxima - Bright Fringes)

If the path difference is an integral multiple of the wavelength $\lambda$, constructive interference takes place at $P$:

$d \sin\theta = m\lambda \quad \text{--- (Maxima)}$

where $\lambda$ is the wavelength of light used and $m$ is the order of the fringe ($m = 0, \pm 1, \pm 2, \pm 3, \dots$). The central bright fringe at $\theta = 0$ ($m = 0$) is called the zero-order maximum.

2. Destructive Interference (Minima - Dark Fringes)

If the path difference is an odd multiple of half-wavelength, the waves arriving at $P$ are out of phase and destructive interference takes place:

$d \sin\theta = \left(m + \frac{1}{2}\right)\lambda \quad \text{--- (Minima)}$

where $m = 0, \pm 1, \pm 2, \pm 3, \dots$

Positions of Bright and Dark Fringes

Since the distance $L$ to the screen is very large compared to the slit separation $d$, the angle $\theta$ is very small ($\sin\theta \approx \tan\theta = \frac{Y}{L}$ from triangle $DOQP$). Thus, the path difference can be expressed as:

$\text{Path Difference} = \frac{d Y}{L}$

Position of Bright Fringes:

$\frac{d Y}{L} = m\lambda \implies Y_m = \frac{m\lambda L}{d}$

Substituting $m = 0, 1, 2, \dots$, the positions of bright fringes are $Y = 0, \frac{\lambda L}{d}, \frac{2\lambda L}{d}, \dots$

Position of Dark Fringes:

$\frac{d Y}{L} = \left(m + \frac{1}{2}\right)\lambda \implies Y_m = \left(m + \frac{1}{2}\right)\frac{\lambda L}{d}$

Substituting $m = 0, 1, 2, \dots$, the positions of dark fringes are $Y = \frac{\lambda L}{2d}, \frac{3\lambda L}{2d}, \frac{5\lambda L}{2d}, \dots$

Fringe Spacing

“The distance between two successive dark or two successive bright fringes is called fringe spacing ($\Delta y$ or $\Delta x G$).”

The fringe spacing $\Delta y$ is given by the difference between the positions of two adjacent bright fringes (e.g., between the 2nd and 1st bright fringe):

$\Delta y = Y_{m+1} - Y_m = \frac{(m+1)\lambda L}{d} - \frac{m\lambda L}{d}$

$\Delta y = \frac{\lambda L}{d}$

If $L$, $d$, and $\Delta y$ are known, the wavelength $\lambda$ of the light can be precisely calculated. The fringe spacing is directly proportional to the wavelength of light and the screen distance, and inversely proportional to the slit separation.

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