Waves: Topic 9.3: Interference

Question 1

What does a double-slit setup result in and why?

Answer 1

Since both light sources originate from the same primary source (monochromatic light), they are coherent and therefore Both diffracted light from the double slits create an interference pattern made up of bright and dark fringes

Question 2

For two-source interference fringes to be observed, the sources of the wave must be:

Answer 2

Coherent (constant phase difference)
Monochromatic (single wavelength)

Question 3

Describe the intensity seen in double slit experiment

Answer 3

For a double-slit interference pattern the intensity of the light is the same for all maxima

Question 4

What is modulation

Answer 4

The resulting interference pattern is a combination of the double-slit and single-slit interference patterns is known as modulation. This is because of the interference of the two diffracted beams, that work on the condition of single-slit diffraction, which requires a significant slit width and the distance between the two slits is much greater than the width as assumptions and conditions.

Question 5

Explain the reason behind the pattern of fringes observed in double-slit interference

Answer 5

The fringes due to the double slits are much closer together than in the single slit case
This is because the distance between the slits is greater than their widths

Question 6

In multiple-slit diffraction, what happens with increasing slit and importance as well as number of secondary maxima

Answer 6

The number of secondary maxima appears in between the maxima (bright fringes) following a pattern of N-2, where N refers to number of slits. With the increasing number of slits, the secondary maxima become unimportant.

Question 7

In multiple-slit diffraction, what happens to width of primary maxima and intensity?

Answer 7

With an increasing number of slits, the width of primary maxima decreases; the bright fringes become very sharp and easily identifiable (width is proportional to 1/N), and intensity N square I0
Where I0 is the intensity of the central maximum of a single slit diffraction pattern

Question 8

The diffraction grating equation is given by:

Answer 8

n lambda equals d sin theta

n = the order of the diffraction pattern
λ = the wavelength of the laser light (m)
d = the distance between the slits (m)
θ = the angle between the normal and the maxima

Question 9

Spacing between each slit, d, can be calculated from N using the equation in diffraction grating

Answer 9

d= 1/N

Question 10

What is angular separation and how to calculate it

Answer 10

The angular separation of each maxima is calculated by rearranging the grating equation to make θ the subject
The angle θ is taken from the centre meaning the higher orders are at greater angles.
The angular separation between two angles is found by subtracting the smaller angle from the larger one
The angular separation between the first and second maxima n1 and n2 is θ2 – θ1

Question 11

The maximum angle to see orders of maxima is when

Answer 11

The maximum angle to see orders of maxima is when the beam is at right angles to the diffraction grating
This means θ = 90 degrees and sin θ = 1

Question 12

The highest order of maxima visible is therefore calculated by the equation

Answer 12

n = d/lambda

Question 13

Define the phenomenon of thin film interference

Answer 13

This phenomenon occurs when light waves reflecting off the top and bottom surfaces of a thin film interfere with one another

Question 14

Conditions for Thin Film Interference

Answer 14

To see the interference light must be incident on a material which:
Is very thin
Has a higher refractive index than the medium surrounding it
Also transmits light
The effect is caused by the reflection of waves from the top and bottom surfaces of the thin film

Question 15

What’s the degree of phase change that happens in thin film interference and where

Answer 15

Part of the light wave reflects at a boundary between a less-dense and a more-dense medium (e.g. Air to Oil where n air < n oil ), and a phase change is seen, such that:
A wave reflected at a boundary with a medium of a higher refractive index undergoes a phase change of half a wavelength (lambda over 2), 180° or π rad

Question 16

Explain the effect of phase change on reflected ray in thin film interference

Answer 16

A wave incident on the boundary with a more dense material is both reflected and transmitted. The transmitted wave continues without a phase change and the reflected wave has a 180° phase change. What was the peak on the incident wave becomes the trough on the reflected wave.

Question 17

Conditions for reflected and transmitted rays in thin film interference with examples

Answer 17

Light is reflected and transmitted from a boundary with a less dense to a more dense material
Air to Water
Water to Oil
Light is transmitted only from a boundary with a more dense to less dense material
Oil to Air

Question 18

Explain what happens. to Light transmitting through the thin film (B)

Answer 18

Part of the light wave also refracts as it enters the thin film and is transmitted through the material to the bottom surface
The light wave is now travelling through a denser medium, so it travels at a slower speed and has a shorter wavelength

Question 19

Explain what happens when Light incident on the bottom surface of the thin film (C) which has a lesser refractive index compared to the next medium

Answer 19

If the bottom surface is at a boundary with a more dense material, then both of the following will occur:
Reflection with a π phase change back into the thin film
Refraction of the transmitted light into the next medium
This is the same situation as the light wave incident on the top surface of thin film

Question 20

Explain what happens when Light incident on the bottom surface of the thin film (C) which has more refractive index compared to the next medium

Answer 20

If the bottom surface is at a boundary with a less dense material (a lower refractive index e.g. oil-air where nair < noil ) then the following will occur:
Transmission with refraction will occur out of the material
No reflection will occur at the boundary