Chapter 8

1

Chapter 8: Physical Optics

8.1 Huygens’ Principle

L.O 8.1.1 State Huygens’ principle

L.O 8.1.2 Sketch and explain the wave front of light after passing through a single slit

and obstacle using Huygens’ principle

Huygens’ Principle:

Every point on a wavefront can be considered as a source of secondary wavelets that spread out

in the forward direction at the speed of the wave. The new wavefront is the envelope of all the

secondary wavelets - i.e. the tangent to all of them.

Plane wave Spherical wave Single slit Obstacle

When applying Huygens’s principle, show

the centres of the wavelets

the wavelets from these centres

the line touching these wavelets

draw an arrow to show the direction of the ray (normal to wavefront)

Example

Question Solution

The figure shows a point light source P on the

ground.

Draw the wavefront from point P at time t = 1

s and t = 2 s.

Wavefront at

time t = 0

New wavefront

at time t

P

Chapter 8

2

8.2 Constructive Interference and Destructive Interference

L.O 8.2.1 Define coherence

A stable interference pattern can be produced if the sources of wave are coherent.

The two sources of wave are coherent if they have:

i. the same phase difference (constant)

ii. the same wavelength (monochromatic)

L.O 8.2.2 State the conditions for interference of light

Interference

When two or more light waves meet at a point, a bright or a dark region will be produced

in accordance to the Principle of Superposition.

Principle of Superposition

The resultant displacement at any point is the vector sum of the displacements due to the

two light waves.

Conditions for fixed interference

two coherent sources

same or approximately same amplitude

distance between the coherent sources, d ≤ λ

L.O 8.2.3 State the conditions of constructive and destructive interference

Constructive interference is

defined as a reinforcement of

amplitudes of light waves that

will produce a bright fringe

(maximum).

Destructive interference is

defined as a total cancellation of

amplitudes of light waves that will

produce a dark fringe (minimum).

Chapter 8

3

Path difference is the difference between two paths

of waves from two different sources at a point (a

difference in path length).

Conditions of constructive and destructive interference (based on path difference):

S1 and S2 are two coherent sources in phase

Bright fringe (Constructive) Dark fringe (Destructive)

Path Difference, L

= |S2P - S1P|

= |x2 –x1|

,.....3,2,1,0

,.....3,2,,0

mwhere

m

L

,.....3,2,1,0

2

1

,.....2

5,

2

3,

2

mwhere

m

L

Chapter 8

4

S1 and S2 are two coherent sources in phase

Bright fringe (Constructive) Dark fringe (Destructive)

Example

Question Solution

Two point sources X and Y emit waves of

wavelength 2.0 cm in phase. The point P is 6.0

cm from X and 10.0 cm from Y. Another point

Q is 7.0 cm from X and 4.0 cm from Y. What

is the path difference of waves from X and Y

at

a. The point P and

b. The point Q?

Hence deduce whether constructive

interference or destructive interference occurs

at P and Q.

,.....3,2,1,0

,.....3,2,,0

mwhere

m

L

,.....3,2,1,0

2

1

,.....2

5,

2

3,

2

mwhere

m

L

Chapter 8

5

8.3 Interference of Transmitted Light through Double-Slits

L.O 8.3.1 Derive and use 𝒚𝒎 =𝒎𝝀𝑫

𝒅 for bright fringes and 𝒚𝒎 =

(𝒎+𝟏 𝟐⁄ )𝝀𝑫

𝒅 for dark

fringes where m = 0, ±1, ±2,±3, …

L.O 8.3.2 Use ∆𝒚 =𝝀𝑫

𝒅 and explain the effect of changing any of the variables

Double-slits interference: Young’s Double Slits Experiment

The double slits, S1 and S2, act as coherent sources of light waves.

Under these conditions, the waves emerging from S1 and S2 have the same frequency

and amplitude and are in phase because they originate from the same light source

and their distance from S is equal.

Constructive interference and destructive interference occur at different points on

the screen to produce a pattern of alternating bright and dark fringes.

Chapter 8

6

Derivation of Young’s double-slit equations

From the figure:

Big triangle: Small triangle:

D

ymtan d

Lsin

since θ is very small, tan θ ≈ sin θ , hence

d

L

D

ym

For bright fringes, mL For dark fringes,

2

1mL

**m = 0, ±1, ±2, ...

Separation between two consecutive (successive) dark or bright fringes:

y depends on :

i) the wavelength of light, λ y

ii) the distance apart, d of the double slits

dy

1

iii) distance between slits and the screen, D dy

d

Dmym

d

Dm

ym

2

1

mm yyy 1

d

Dm

d

Dmy

1

d

Dy

Chapter 8

7

Example

Question Solution

An interference pattern is formed on a screen

when light of wavelength 550 nm is incident

on two parallel slits 50 μm apart. The second-

order bright fringe is 4.5 cm from the center

of the central maximum. How far from the

slits is the screen?

In a Young’s double experiment, the slits

separation is 1.0 mm. The distance between

the slits and the screen is 1.0 m. The

wavelength of the sodium light used is

5.9×10-5 cm.

a. Calculate the separation between two

consecutive dark fringes.

b. If the sodium light is replaced with a blue

light, what are the changes to the

interference pattern on the screen?

A double-slits pattern is view on a screen

1.00 m from the slits. If the third order

minima are 25.0 cm apart, determine the

distance between the first order minimum and

fourth order maximum on the screen.

A monochromatic light of wavelength 600

nm falls on a system of double-slits of

unknown slit separation. At the same time,

the double-slits is illuminated by a

monochromatic light of unknown

wavelength. It was observed that the 4th order

maximum of the known wavelength light

overlapped with the 5th order maximum of the

unknown wavelength light. Find the

wavelength of the unknown wavelength light.

Chapter 8

8

Question Solution

Suppose you pass the light from a He-Ne

laser through two slits separated by 0.0100

mm and find that the third bright line is

formed at an angle of 10.95o relative to the

incident beam. What is the wavelength of the

light?

Exercise

Question

In a lab experiment, monochromatic light passes through two narrow slits that are 0.050 mm

apart. The interference pattern is observed on a white wall 1.0 m from the slits, and the

second-order bright is 2.4 cm from the center of the central maximum.

a. What is the wavelength of the light?

b. What is the distance between the second-order and third-order bright fringes?

Answer: 6.7×10-7 m, 1.2×10-2 m

Monochromatic light illuminates a double-slit system with a slit separation d = 0.30 mm. The

second-order maximum occurs at y = 4.0 mm on a screen 1.0 m from the slits. Find

a. the wavelength.

b. the distance y on the screen between the central maximum and the third order-minimum.

c. the angle, θ of the first-order maximum.

Answer: 600 nm, 7mm, 0.11°

A screen containing two slits 0.100mm apart is 1.20 m from the viewing screen. Light of

wavelength λ = 500 nm falls on the slits from a distant source.

a. Approximately how far apart will the bright interference fringes be on the screen?

b. What happens to the interference pattern if the incident light (500 nm) if replaced by light

of wavelength 700 nm.

c. What happens instead if the slits are moved farther apart?

Answer: 6 mm, u think

What is the highest-order maximum for 400 nm light falling on double slits separated by 25.0

μm?

Answer: 62

Chapter 8

9

8.4 Interference of Reflected Light in Thin Films

L.O 8.4.1 Identify the occurrence of phase change upon reflection

L.O 8.4.2 Explain with the aid of a diagram the interference of light in thin films at

normal incidence

L.O 8.4.3 Use 𝟐𝒏𝒕 = 𝒎𝝀 and 𝟐𝒏𝒕 = (𝒎 + 𝟏𝟐⁄ )𝝀

Less dense →Denser Denser →Less dense

Phase change = π rad Phase change = 0 rad

Non-reflective coating Reflective coating

Phase difference between ray 1 and ray 2,

∆𝜙 = 𝜋 − 𝜋 = 0 2 sources in phase

Path difference between ray 1 and ray 2,

∆𝐿 = 2𝑛𝑡

Constructive interference:

mnt 2

Destructive interference:

2

12 mnt

Phase difference between ray 1 and ray 2,

∆𝜙 = 𝜋 − 0 = 𝜋 2 sources anti phase

Path difference between ray 1 and ray 2,

∆𝐿 = 2𝑛𝑡

Constructive interference:

2

12 mnt

Destructive interference:

mnt 2

**m = 0, ±1, ±2, ...

Reflected Reflected Transmitted Transmitted

Chapter 8

10

Example

Question Solution

Light is at normal incidence on a thin soap

film of refractive index 1.30 and thickness

0.15 µm. Determine the maximum

wavelength of the reflected light that

undergoes

a. constructive interference

b. destructive interference

White light is incident normally on a lens

(n = 1.52) that is coated with a film of MgF2

(n = 1.38). For what minimum thickness of

the film will yellow light (λvacuum = 550 nm)

be missing in the reflected light?

A lens appears greenish yellow (λvacuum = 570

nm is strongest) when white light reflects

from it. What minimum thickness of coating

(n = 1.25) do you think is used on such a

(glass) lens?

Exercise

Question

White light is incident on a soap film of refractive index 1.30 in air. The reflected light looks

bluish because the red light of wavelength 670 nm is absent in the reflection. What is the

minimum thickness of the soap film?

Answer: 2.58×10-7 m

A non-reflective coating of MgF2 (n = 1.38) covers the glass (n = 1.52) of a camera lens.

Assuming that the coating prevents reflection of yellow-green light (λvacuum = 565 nm) ,

determine the minimum nonzero thickness that the coating can have.

Answer: 102 nm

A plastic film with index of refraction 1.80 is put on the surface of a car window to increase

the reflectivity and thereby to keep the interior of the car cooler. The window has index of

refraction 1.60. What minimum thickness required if light of wavelength 600 nm in air

reflected from the two sides of the film is to interfere constructively.

Answer: 83 nm

Chapter 8

11

8.5 Interference of Reflected Light in Air Wedge and Newton’s Rings

L.O 8.5.1 Explain with the aid of a diagram the interference of light in air wedge

L.O 8.5.2 Use 𝟐𝒕 = 𝒎𝝀 for dark fringes and 𝟐𝒕 = (𝒎+ 𝟏𝟐⁄ )𝝀 for bright fringes

L.O 8.5.3 Use diagram to explain qualitatively the formation of Newton’s rings and

the centre dark spot

Air wedge Newton’s rings

˃ Ray 1 (OL): From denser to less dense, no phase change

˃ Ray 2 (BQ): From less dense to denser, π rad phase change

˃ The two rays (ray 1 & 2) are coherent since both have originated from the

same source and produces a produces interference pattern. When thickness

is equal to zero, destructive interference occurred. Hence a dark fringe (Air

wedge)/ central dark spot (Newton’s ring) is obtained.

Equations for air wedge:

For bright fringes, For dark fringes,

2

12 mnt mnt 2

**m = 0, 1, 2, ...

Chapter 8

12

Example

Question Solution

An air wedge is formed by placing a human

hair between two glass slides of length 44

mm on one end, and allowing them to touch

on the other end. When this wedge is

illuminated by a red light of wavelength

771 nm, it is observed to have 265 bright

fringes. Determine

a. the diameter of hair

b. the angle of air wedge

c. the thickness of the air film for 99th dark

fringe to be observed

d. the separation between two consecutive

bright fringes

A plate of glass 10.0 cm long is placed in

contact with a second plate and held at small

angle with it by a metal strip 0.100 mm thick

placed under one end. The space between the

plates is filled with air. The glass is

illuminated from above with light having a

wavelength of 635 nm. How many

interference fringes are observed per

centimeter in the reflected light?

Exercise

Question

A fine metal foil separates one end of two pieces of optically flat glass. When light of

wavelength 670 nm is incident normally, 28 dark lines are observed (with one at each).

a. How thick is the foil?

b. How far apart are the dark fringes if the glass plates are each 26.5 m long?

Answer: 9.05×10-6 m, 9.8 mm

White light is incident normally on a thin soap film (n =1.33) suspended in air.

a. What are the two minimum thicknesses that will constructively reflect yellow light of

wavelength 590 nm?

b. What are the two minimum thicknesses that will destructively reflect yellow light of

wavelength 590 nm?

Answer: 110 nm, 330 nm, 220 nm, 440 nm

Chapter 8

13

8.6 Diffraction by a Single Slit

L.O 8.6.1 Define diffraction

L.O 8.6.2 Explain with the aid of a diagram the diffraction of a single slit

L.O 8.6.3 Use 𝒚𝒏 =𝒏𝝀𝑫

𝒅 for dark fringes and 𝒚𝒏 =

(𝒏+𝟏 𝟐⁄ )𝝀𝑫

𝒅 for bright

fringes where n = ±1, ±2,±3, …

Diffraction of light is defined as the bending of waves as they travel around obstacles or pass

through an aperture comparable to the wavelength of the waves.

Diffraction by a single slit

Formation of diffraction by a single slit:

According to Huygen’s principle, wavefront from light

source falls on a narrow slit and diffraction occurs. Every

point on the wavefront that falls on the slit acts as sources

of secondary wavelets and superposed each another to

form an interference pattern on the screen.

Chapter 8

14

For bright fringes, For dark fringes,

**n = ±1, ±2, ...

Example

Question Solution

A monochromatic light of wavelength 6 x10-7

m passes through a single slit of width 2 x 10-

6 m.

a. Calculate the width of central maximum:

i. in degrees;

ii. in centimeters,

on a screen 5 cm away from the slit

b. Find the number of minimum that can be

observed.

How many bright fringes will be produced on

the screen if a green light of wavelength 553

nm is incident on a slit of width 8.00 µm?

Exercise

Question

Visible light of wavelength of 550 nm falls on a single slit and produces its second diffraction

minimum at an angle of 45.0o relative to the incident direction of the light.

a) What is the width of the slit?

b) At what angle is the first minimum produced?

Answer: 1.56×10-6 m, 20.7°

Important!!

» The width of central maximum is

2y1 (equation for first dark)

» To calculate the maximum number

of orders observed, θ = 90o

nn sina

2

1sina nn

a

Dnyn

)(21

a

Dnyn

Chapter 8

15

8.7 Diffraction Grating

L.O 8.7.1 Explain with the aid of a diagram the formation of diffraction

L.O 8.7.2 Apply 𝒅𝐬𝐢𝐧 𝜽 = 𝒏𝝀 where 𝒅 =𝟏

𝑵

Diffraction grating is defined as a large number of equally spaced parallel slits.

Pattern of diffraction grating:

Formation of diffraction:

A diffraction grating is a plate containing many parallel lines/slits at uniform distance between

one another. According to Huygen’s principle, when light is incident on a diffraction grating,

each slit will become a secondary source of light so that superposition of light waves from each

source will produce diffraction images of regular orders on a screen

Path difference

3,...2,1,0,n

sin

nd n

Nd

1

Slit separation Number of lines

per unit length

Chapter 8

16

Example

Question Solution

A diffraction grating with 600 lines per mm is

illuminated normally with a monochromatic

light of wavelength 589 nm. Calculate

a. the angles of the first-order and second-

order maximum lines from the zero-order

maximum line.

b. the number of orders that can be

observed.

When a blue light of wavelength 465 nm

illuminates a diffraction grating, it produces a

1st order maximum but no 2nd order

maximum.

a. Explain the absence of 2nd order

maximum.

b. What is the maximum spacing between

lines on this grating?

How many bright fringes are produced when

a grating with a spacing of 2.00 x 10-6 m is

illuminated normally with light of

wavelength 6.44 x 10-7 m?

Exercise

Question

A monochromatic light of unknown wavelength falls normally on a diffraction grating. The

diffraction grating has 3000 lines per cm.

If the angular separation between the first order maxima is 35. Calculate

a. the wavelength of the light,

a) the angular separation between the second-order and third- order maxima.

Answer: 1.00×10-6 m, 27.4°

The second-order maximum produced by a diffraction grating with 560 lines per centimeter

is at an angle of 3.1.

a. What is the wavelength of the light that illuminates the grating?

b. Determine the number of maximum can be observed on a screen.

c. State and giving reason, what you would expect to observe if a grating with a larger

number of lines per centimeter is used.

Answer: 4.83×10-7 m, 37, u think