AH 3.3.5 Interference by division of amplitude

Question 1

Look at the diagram, which shows two rays of light, a and b.

a) What is meant by the geometric path length l, from source to observer for each ray?
b) Write down the expression for the optical path length (OPL) of each ray from the beginning to the end of the glass block.
c) Write down an expression for the optical path difference between the two rays.
c) Write down an equation which applies when the two rays interfere constructively for the observer.
e) Write down an equation which applies when the two rays interfere destructively for the observer.

Answer 1

a) The geometric path length l is simply the length in metres between source and observer.
b) Ray a: optical path length OPL = n_a d

Ray b: optical path length OPL = n_g d

c) The optical path difference = n_gd- n_ad = (n_g- n_a) d
d) Constructive interference: (n_g- n_a) d = mλ m = 0,1,2,3 ….
e) Destructive interference: (n_g- n_a) d = (m + ½) λ

Question 2

A hollow air filled perspex microfibre is shown below. Light of wavelength 700 nm passes through and around the microfibre.

a) Determine the optical path length between A & B.
b) A ray of light follows the path AB. Another ray follows the path CD, just outside the block. What is the optical path difference between the two rays?
c) What is the phase difference between the two rays (in radians)?

Answer 2

a) Perspex: n = 1.50

A → B: OPL = nd₁ + d₂ + nd₃
= (1.50 x 40 x10^-9) + (70 x 10^-9) + (1.50 x 40 x 10^-9)
= 190 nm

b) C → D: OPL = 150 nm
So OPD = 190 - 150
= 40 nm

c) In general, phase difference δ = (2π/ λ) x where x = separation in metres

So, here δ = (2π/ λ) x OPD
= (2π / 700 x 10^-9) x 40 x 10^-9
= 0.36 radians

Question 3

The diagram shows a ray of light in air incident upon a thin layer of oil which sits upon a base layer of water. The respective refractive indices are shown.

What phase change, if any, takes place at:

a) the air-oil boundary?
b) the oil-water boundary
c) explain your answers to (a) and (b)
d) Will the two reflected rays be coherent?
e) Explain your answer to (d)

Answer 3

a) π phase change
b) No phase change
c) When light passes into a more optically dense medium there is a partial reflection and the reflected ray undergoes a π phase change.

When light passes into a less optically dense medium there is a reflection but no phase change.

d) Yes - the two rays will be coherent
e) Because they originate from the same source.

The original ray has been divided in two i.e. its amplitude has been divided.

This is interference by division of amplitude.

Question 4

Look at the diagram below.

a) Write down an expression for the optical path difference between the two reflected rays.
b) Establish an equation for constructive interference between the two reflected rays.
c) Establish an equation for destructive interference between the two reflected rays.

Answer 4

a) optical path difference OPD = 2n_od where n_o = refractive index of the oil
b) Constructive interference 2n_od = (m + ½) λ
c) Destructive interference 2n_od = m λ

Question 5

Look at the diagram below.

What is the phase change at:

a) The first boundary?
b) The second boundary?
c) Explain your answers to (a) and (b)

Answer 5

a) No phase change
b) No phase change
c) In both cases the light ray passes into a less optically dense medium so there is no phase change.

Question 6

The diagram shows the glass of a camera lens which has been coated with a thin layer of Magnesium Flouride (Mg F).

a) What is the purpose of the Mg F coating?
b) What is the process of adding such a coating called?

NOTE THAT PARTS (c) and (d) BELOW TOGETHER CONSTITUTE A FORMAL DERIVATION SO MUST BE MEMORISED

c) Derive the equation for the thickness of coating which results in destructive interference between the two reflected rays.
d) Derive the equation for the minimum thickness of coating which results in destructive interference between the two reflected rays.

Answer 6

a) To reduce reflection of light from the lens and so increase transmission through it.
b) Blooming

DERIVATION:

c) Both rays undergo a π = λ / 2 phase change upon reflection

Destructive interference : 2n_cd = (m+ ½ ) λ
d = (m+ ½ ) λ / 2n_c

d) For d to be minimum, m must be minimum so m=0

Hence d_min = ½ λ / 2n_c

d_min = λ / 4n_c

Question 7

a) What thickness of coating is required to give non-reflection in green light of wavelength 540 nm for a lens of refractive index 1.53.
b) Explain why lenses with a non-reflective coating appear to have a purple hue when white light is incident.

Answer 7

a) λ = d / 4n

d = 4nλ = 4 x 1.53 x 540 x 10^-9

d = 3.30 x 10^-6 m

b) The non-reflective coating only causes cancellation for a specific wavelength, λ, where λ = d/4n.

The chosen λ is normally that of green light which has the average λ for the visible spectrum (around 550 nm). Such cancellation of green means that red and violet dominate and combine to give a purplish appearance.

Question 8

The diagram shows two glass microscope slides separated at one end so as to form an air wedge.

a) Explain how wedge fringes are created by this arrangement.

The slides are 100 mm long. They are separated by a piece of paper 30 μm thick, placed at ‘y’ in the diagram. Light of wavelength 650 nm is used. The interference pattern is viewed through a travelling microscope.

b) Sketch the fringe patter observed.
c) Calculate the separation of the fringes.

Answer 8

a) Incident light reflects at the first glass-air boundary and also at the lower air-glass boundary. The reflected rays interfere constructively and destructively to produce bright and dark fringes.
b) see the diagram below.
c) Δx = λl/2d where l = thickness of paper called ‘y’ in diagram

= 650 x 10^-9 x 100 x 10^-3 / 2 x 30 x 10^-6

Δx = 1.1 x 10^-3 m

Fringe pattern observed