# Interference

## Interference Revision

**Interference**

We have discussed **interference** in previous pages but this section will look into the process of interference in much more detail. We will look in depth at the process of **interference** and some experiments that show the effect of **interference**.

**Coherence**

**Coherent waves** are often used when observing **interference**. Two waves are coherent if they have the same **frequency**, same **wavelength** and are kept at a constant **phase difference**.

We do not often study **interference** using white light because it is a **spectrum** of colours, each with a different **frequency** and **wavelength**. It is therefore, not **coherent**.

The best source of **coherent** light is a **laser**. A laser produces light of a single **frequency**, **wavelength** and at a constant **phase difference**. When using lasers it is important to consider the safety implications of using the laser:

- Only use the
**recommended power laser**. Schools are only permitted to use lasers of maximum power 1\text{ mW} which minimises the risk to students. - Ensure
**suitable warnings**are placed around the room and outside of the room to make people aware of the danger. - Appropriate
**eye-wear**should be worn and the laser should never be pointed toward someone’s face/eyes. **Stand behind the laser**unless completely necessary to move in front.**Avoid shiny reflective materials**that may reflect the laser as even the reflection may cause permanent damage.

**Path Difference**

The **path difference** of two **waves** can be defined as:

**The difference in the distance travelled by two waves to a given point**.

This is measured in metres and each distance should be measured from the source to the point where the two waves meet, usually in a screen for most experiments. However, this measurement is usually expressed in terms of the **wavelength** of the wave e.g 10 \lambda meaning 10 times the **wavelength**.

If the path difference of two coherent waves is a whole number of **wavelengths** (n \lambda), **constructive interference** occurs. This occurs as the **peaks** and **troughs** of the waves line up and **superposition** occurs.

If the **path difference** is a whole number and a **half ****wavelength** (n+\dfrac{1}{2})\lambda, then complete **destructive interference** occurs as the peaks line up with the troughs of each wave, completely** cancelling** each other out.

**Interference:**** **

When two waves meet and **interfere**, they form a wave with the **amplitude** of the two waves combined.

**Constructive interference** – The diagram above shows **constructive interference**. The two waves on the bottom are **in-phase** and therefore **constructively interfere** producing a wave with the **amplitude** of both waves combined. When the waves are completely **in-phase**, a new wave of **maximum amplitude** is formed.

**Destructive Interference** – the diagram above shows **destructive interference**. The two waves at the bottom are in **anti-phase** and therefore **destructively interfere**. When the two **wavelengths** are combined, they cancel each other out forming a wave with zero **amplitude** throughout.

**Diffraction** occurs any time a wave passes through a slit of past an object comparable to the wavelength of the wave. When the light from the **diffraction grating** or **double slit** is displayed in a screen, a distinctive pattern is shown.

The light fringes displayed in the screen show areas where the waves have hit the screen with a whole number of **wavelengths path difference**. Therefore, **constructive interference** occurs, resulting in a bright **maxima** of light.

The dark fringes displayed in the screen show areas where the waves have hit the screen with a whole number and a half of **wavelengths path difference**. Therefore, **destructive interference** occurs, resulting in a dark **minima** of light.

The central bright fringe is the brightest of the **maxima** and is where there is no **path difference**. We label this n=0. Outwards from the centre in both directions, each bright fringe appears at each time the path difference is a whole number of **wavelengths** out. At n=1, the **path difference** is exactly one wavelength, n=2 is where the path difference is 2 wavelengths etc. The **maxima** gradually get dimmer as you move out from the centre due to the light intensity decreasing with distance

**Fringe Spacing Equation**

The **fringe spacing** equation allows us to calculate the distance between the centre of each bright fringe. The equation is:

w=\dfrac{\lambda D}{s}

- w= the
**fringe spacing**in metres (m) - \lambda= the
**wavelength**in metres (m) - D= the d
**istance between the double slit and the screen**in metres (m) - s= the
**distance between the centre of the two slits**in metres (m).

**Required Practical 2**

**Investigating the relationship between the distance separating two slits and the screen and the fringe width.**

**Doing the experiment:**

- Set up the experiment as shown in the diagram above.
- Set the distance D at a fixed distance such as 1 \text{ m}.
- Ensure the room is dark and
**safety labels**are put around the room before turning the laser on. - Using
**vernier callipers**, measure the fringe spacing of multiple bright fringes then divide by the number of fringes to give an accurate measure of w_1. - Repeat the experiment, increasing the distance D by 0.1 \text{ m} each time.
- Repeat until you have at least 3 sets of results for each value of D

**Analysing the results:**

The results collected can be compared to our expected results using the equation:

w=\dfrac{\lambda D}{s}

- w= the
**fringe spacing**in metres \text{(m)} - \lambda = the
**wavelength of light**in metres \text{(m)} - D = the
**distance between the double slit and the screen**in metres \text{(m)} - s= the
**distance between the centre of the two slits**in metres \text{(m)}

A graph of w against D may be drawn where w is on the y-axis and D is on the x-axis. The graph formed will be a linear graph where the **gradient** is equal to \dfrac{\lambda}{s} . Therefore multiplying the gradient of the graph by the slit spacing (s) will give you the wavelength of light.

We may expect to have a **degree of error** in our measurements of small distances. This can be reduced by taking **multiple readings** and by measuring the fringe spacing over multiple fringes.

**Example: Calculating Fringe Spacing**

A laser of frequency 500 \text{ THz} is directed through a double slit which are spaced 0.5 \text{ mm} apart. If the distance between the double slits and the screen is 3 \text{ m}, what is the fringe spacing?

**[3 marks]**

First, calculate the wavelength:

\begin{aligned} c&=f\lambda \\ \lambda &= \dfrac{c}{f} \\ &= \dfrac{3\times 10^8}{\textcolor{f95d27}{500 \times 10^{12}}} \\ &= \bold{6\times 10^{-7}} \textbf{ m} \end{aligned}

Substitute values into the fringe spacing equation:

\begin{aligned} \bold{w} &= \bold{\dfrac{\lambda D}{s}} \\ &= \dfrac{6\times 10^{-7} \times \textcolor{f21cc2}{3}}{\textcolor{aa57ff}{0.5 \times 10^{-3}}} \\ &= \bold{3.6 \times 10^{-3}} \textbf{ m or } \bold{3.6} \textbf{ mm} \end{aligned}

## Interference Example Questions

**Question 1:** Describe what is meant by coherence.

**[2 marks]**

Coherent waves are waves that have the **same wavelength**** and** **frequency** and have a **constant path difference**.

**Question 2:** Describe how minima and maxima are formed on a screen during double slit diffraction.

**[2 marks]**

Minima are formed where there is a **path difference of a whole number and a half wavelength difference**. **Destructive interference** occurs causing the waves to cancel each other out.

Maxima are formed where there is a **path difference of a whole number of wavelengths difference.** **Constructive interference** occurs causing the waves to undergo superposition and produce a high intensity of light.

**Question 3:** A laser of frequency 700 \text{ THz} is directed through a double slit which are spaced 0.75 \text{ mm} apart. If the distance between the double slits and the screen is 2.3 \text{ m}, what is the fringe spacing?

**[3 marks]**

\begin{aligned} \lambda &= \dfrac{c}{f} \\ &= \dfrac{3 \times 10^8}{700 \times 10^{12}} \\ &= \bold{4.286 \times 10^{-7}} \textbf{ m} \end{aligned} \\ \begin{aligned} \bold{w} &= \bold{\dfrac{\lambda D}{s}} \\ &= \dfrac{4.286 \times 10^{-7} \times 2.3}{0.75 \times 10^{-3}} \\ &= \bold{1.31 \times 10^{-3}} \textbf{ m or } \bold{1.31} \textbf{ mm} \end{aligned}

## Interference Worksheet and Example Questions

### Combining Waves Questions

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