Double slit interferenceBy: Pankhil Kainth

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Thomas young postulated the idea that light is wave and has to obey the rules of the principle of superposition. He used this idea to demonstrate constructive and destructive interference in the 19th century. In the experiment, a light source is illuminates two parallel slits on a mainly opaque surface. Two shadows are created due to the light on the screen. The principle of superposition determines the intensity of the resulting pattern on the screen. Constructive interfernece has taken place whenever the difference in the paths is an integral number of wavelengths. This means that two waves’ crests have arrived at the same time.

THE EXPERIMENT

LIGHT• Properties of light behaving as waves were first

demonstrated by Thomas Young.• In his experiment, a single source of light illuminated

two slits on a screen. The pattern of light passing through the slits was observed on a screen

LIGHT

• When light waves interfere, a pattern of dark and light bands can be observed.

• Electrons also produce these interference patterns because of their similar wave like nature

• The light and dark bands are called interference fringes

• Dark bands are where destructive interference takes place

• Light bands are regions of constructive interference.

INTERFERENCE

• When bright fringes occur, the wave is in phase

• If l is the angle of the beams from the normal

• The distance between the two slits is d, forming two triangles, for dark and bright fringes

EQUATION

Using the triangles from the previous slide we can figure out:

Then,

Bright fringes can occur at multiple wavelengths, as long as it’s an integer

Where n can be any integer

Destructive Interference

In a similar way, dark bands are calculated like bright fringes except an extra distance Is added which is equal to the odd number of half wavelengths:

d sin = (q m+1/2)l

“m” denotes how many wavelengths fit in the extra distance• If the distance from the screen L is greater than d, than the

thickness of the bands on the screen will greatly exceed the wavelength lambda.

• Therefore, given a large ratio of L/d it is possible to observe a pattern, despite the fact that the visible light wavelength is of the order 1 micron.

Problem 1: A double slit interference is being created when two slits are placed 0.27nm apart. The distance between the fifth and first minimum on a screen which is 40cm behind the slits is 6mm.

What is the wavelength of the in nm used in this particular experiment?

Solution:

We know that d=0.27mmThe Δy=6mm,

Then the L=40cm=400mmTherefore, d*(Δy/L)= 4lambdaThis is the distance between the first and the fifth minimum

0.27*(6/400) =4 lambdaThen, lambda=1.01*10^-3mm

Which is 1010nm

Does the distance to the first minima in the diffraction envelope

A)IncreaseB)DecreaseC)Stay the same

1.When the slit separation is increased?2. When the slit width is increased?

1. B=hD/dWhere h is the wavelength of the light source usedD is the distance between the slit sources and the screend is the distance between the two slits

By looking at the equation we see that, when the slit distance d is increased, the fringe width B decreases

Therefore the distance to the first minima will decrease

2. Then when the slit width is increased, there is no effect on the fringe width which causes no change in the difference to the distance to the first minima.

However, the pattern of interference does lose the clarity and intensity as each maxima reduces.