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Interference of Light



Coherent And Incoherent Addition Of Waves

When two or more waves overlap in a region of space, their effects add up according to the principle of superposition. The nature of the resulting wave depends significantly on the relationship between the phases of the individual waves. This leads to the concepts of coherent and incoherent addition of waves.


Coherent Sources

Two sources of waves are said to be coherent if they produce waves of the same frequency and have a constant phase difference between them over time. If the phase difference is always zero, they are in phase. If the phase difference is constant but non-zero, they maintain that relative phase over time.

Coherent sources are required to observe sustained interference patterns. In a coherent wave, the phase relationship between different points in space or different points in time is predictable and constant.

Examples of coherent sources (or how to obtain them):


Incoherent Sources

Two sources of waves are said to be incoherent if they produce waves whose phase difference varies randomly and rapidly with time. For example, light from a conventional bulb consists of waves emitted by individual atoms, which radiate light pulses randomly and independently. The phase relationship between the waves from different atoms (or even different pulses from the same atom) changes rapidly and unpredictably.

If two incoherent sources are used, while interference occurs instantaneously according to superposition, the pattern averages out over time due to the rapid random phase changes. No stable interference pattern is observed.


Addition of Waves

Interference phenomena are observed only with coherent sources because a stable phase relationship is needed for the points of constructive and destructive interference to remain fixed over time, creating an observable pattern of varying intensity.



Interference Of Light Waves And Young’S Experiment ($ \beta = \frac{\lambda D}{d} $)

Interference of light is the phenomenon that occurs when two or more light waves superpose, resulting in a redistribution of light intensity in the region of overlap. This redistribution creates an interference pattern, which typically consists of alternating bright and dark fringes.

Interference is a strong evidence for the wave nature of light. It can only occur if the interfering waves are coherent.


Young's Double-Slit Experiment (YDSE)

Thomas Young performed his famous double-slit experiment in 1801, which provided compelling evidence for the wave theory of light. In this experiment, monochromatic light (light of a single wavelength or frequency) from a single source (S) is passed through two narrow, closely spaced parallel slits ($S_1$ and $S_2$) in an opaque screen. These two slits act as two coherent secondary sources of light waves because they are illuminated by the same wavefront from the original source.

Diagram of Young's Double-Slit Experiment showing path difference and interference pattern.

(Image Placeholder: A diagram showing a single slit illuminated by a source. A screen with two narrow parallel slits S1 and S2 is placed behind the first slit. A screen is placed some distance away from the double slit. Show waves from S1 and S2 spreading out and overlapping on the screen. Show points of constructive (bright fringes) and destructive (dark fringes) interference on the screen, forming alternating bands.)

The waves from $S_1$ and $S_2$ are coherent and travel to a screen placed at a distance $D$ from the double slit. At any point P on the screen, the waves from $S_1$ and $S_2$ interfere. The outcome of the interference at point P (constructive or destructive) depends on the path difference between the waves travelling from $S_1$ to P and from $S_2$ to P, i.e., $\Delta x = S_2P - S_1P$.

Assuming the distance $D$ to the screen is much larger than the distance between the slits $d$ ($D \gg d$), and the point P is at a distance $y$ from the center of the screen (O), the path difference $\Delta x$ can be approximated as:

$ \Delta x = S_2P - S_1P \approx \frac{yd}{D} $ (for small angles, assuming P is close to the central axis)


Conditions for Constructive and Destructive Interference


Position of Bright and Dark Fringes

Using the approximation $\Delta x \approx yd/D$, we can find the positions of the fringes on the screen:


Fringe Width ($\beta$)

The distance between two consecutive bright fringes or two consecutive dark fringes is called the fringe width ($\beta$) or bandwidth. For bright fringes: $\beta = y_{n+1}^{bright} - y_n^{bright} = \frac{(n+1)\lambda D}{d} - \frac{n\lambda D}{d} = \frac{\lambda D}{d}$. For dark fringes: $\beta = y_{n+1}^{dark} - y_n^{dark} = (n+1+\frac{1}{2})\frac{\lambda D}{d} - (n+\frac{1}{2})\frac{\lambda D}{d} = \frac{\lambda D}{d}$.

The formula for fringe width is:

$ \beta = \frac{\lambda D}{d} $

where:

This formula shows that the fringe width is directly proportional to the wavelength and the screen distance, and inversely proportional to the slit separation. By measuring the fringe width, the wavelength of light can be determined.

Young's experiment provided compelling evidence for the wave nature of light and allowed for the first determination of the wavelengths of visible light.

Example 1. In Young's double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is 1.2 cm. Calculate the wavelength of the light used.

Answer:

Slit separation, $d = 0.28$ mm $= 0.28 \times 10^{-3}$ m.

Distance to screen, $D = 1.4$ m.

Distance between the central bright fringe ($n=0$) and the fourth bright fringe ($n=4$) is 1.2 cm. This distance is $y_4^{bright} - y_0^{bright} = 1.2$ cm $= 1.2 \times 10^{-2}$ m.

Position of the $n$-th bright fringe from the central fringe is $y_n^{bright} = \frac{n\lambda D}{d}$.

So, $y_4^{bright} = \frac{4\lambda D}{d}$. Since $y_0^{bright} = 0$, $y_4^{bright} - y_0^{bright} = y_4^{bright}$.

$ 1.2 \times 10^{-2} \text{ m} = \frac{4 \times \lambda \times 1.4 \text{ m}}{0.28 \times 10^{-3} \text{ m}} $

Rearrange to solve for $\lambda$:

$ \lambda = \frac{(1.2 \times 10^{-2} \text{ m}) \times (0.28 \times 10^{-3} \text{ m})}{4 \times 1.4 \text{ m}} $

$ \lambda = \frac{1.2 \times 0.28 \times 10^{-5}}{5.6} $ metres.

$ \lambda = \frac{0.336 \times 10^{-5}}{5.6} = 0.06 \times 10^{-5} $ metres.

$ \lambda = 6 \times 10^{-2} \times 10^{-5} = 6 \times 10^{-7} $ metres.

Convert to nanometres (1 nm $= 10^{-9}$ m): $\lambda = 6 \times 10^{-7} \times 10^9 \times 10^{-9}$ m $= 600 \times 10^{-9}$ m $= 600$ nm.

The wavelength of the light used is $6 \times 10^{-7}$ metres, or 600 nm. This falls in the orange-red part of the visible spectrum.

Example 2. In YDSE, if the distance between the slits is halved and the distance between the slits and the screen is doubled, what happens to the fringe width?

Answer:

The formula for fringe width is $\beta = \frac{\lambda D}{d}$.

Let the original slit separation be $d_1$ and the original screen distance be $D_1$. The original fringe width is $\beta_1 = \frac{\lambda D_1}{d_1}$.

The new slit separation is $d_2 = d_1/2$.

The new screen distance is $D_2 = 2D_1$.

The wavelength $\lambda$ remains unchanged (assuming the same light source). The new fringe width is $\beta_2 = \frac{\lambda D_2}{d_2}$.

Substitute the new values in terms of the original values:

$ \beta_2 = \frac{\lambda (2D_1)}{(d_1/2)} $

$ \beta_2 = \frac{2\lambda D_1}{(d_1/2)} = 2\lambda D_1 \times \frac{2}{d_1} $

$ \beta_2 = 4 \frac{\lambda D_1}{d_1} $

Since $\beta_1 = \frac{\lambda D_1}{d_1}$, we have:

$ \beta_2 = 4 \beta_1 $

The new fringe width is four times the original fringe width. The fringes will be wider and more spread out on the screen.