Module NF2FSLT SP025

Non Face-to-face SLT

Exercise 145

Solution Path difference, PD = 13/4 - 11/4= 1/2

Non Face-to-face SLT

Condition Constructive Interference

•The wave from Source 1 travels a distance of 2¼ λ.
• The wave from Source 2 travels a distance of 3¼ λ to reach the same point, P.
•The difference in distance traveled by the two waves is one full wavelengths;

thus PD is λ. When the PD is one full wavelengths, a crest meets a crest and
constructive interference occurs.

PATH DIFFERENCE = zero or whole number of λ

Non Face-to-face SLT

Condition Destructive Interference

•Wave from Source 1 travels a distance of 2¾λ.

•The wave from Source 2 travels a distance of 3¼ λ to reach the same point.

•The difference in distance traveled by the two waves is one-half of a
wavelength; thus PD is ½λ . When the path difference is one-half a
wavelength, a crest meets a trough and destructive interference occurs.

PATH DIFFERENCE = ½ λ

But for ANTI-PHASE coherent sources, the condition are
reversed.

Non Face-to-face SLT

Exercise 146

1. What are the necessary conditions for the
interference of light to be observable?

(1)The light sources must be coherent ( maintain constant phase
difference).

(2)The light sources must have identical wavelength
(monochromatic).

(3)The light sources must have equal or approximately equal
amplitudes.

Exercise 147
2. Explain what is meant by the term path difference with
reference to the interference of 2 wave motions.

Path difference is difference in distance traveled by two light
waves from their respective sources to the same point in a region.

Exercise 148 Non Face-to-face SLT

3. Why is it not possible to see interference where the
light beams from the headlamps of a car overlap?

Light beams from the headlamps are incoherent.

Exercise 149

4. What do you understand by (i) interference, (ii) coherence
between two separate wave trains?

(i) Interference occurs whenever the two wave trains
overlap at a given point.

(ii) Coherence means both wave trains maintain a
constant phase difference.

Non Face-to-face SLT

Exercise 150

A viewing screen is separated from a double-slit source by 1.2
m. The distance between the two slits is 0.030 mm. The
second bright fringe is 4.5 cm from the center line.
(a) Determine the wavelength of the light.
(b) Calculate the distance between adjacent bright fringes.

Solution

Given: D  1.2 m Y2
D
with d  0.03103 m m d
m  2, Y 2  4.5102

Non Face-to-face SLT

  dYm

mD

 0.03103 (4.5 102 )
(2)1.2

  5.6107 m

(b) Using:

y   D  560109 (1.2)
0.03103
d

 2.2 102 m

Non Face-to-face SLT

Exercise 151

In a Young’s double slit experiment, the slit separation is 0.05 cm
and the distance of the double slit from the screen is 200 cm. When
blue light is used, the distance of the first bright fringe from the
centre of the interference pattern is 0.13 cm.
(a) Calculate the wavelength of the blue light.
(b) Calculate the distance of the fourth dark fringe from the centre

of the interference pattern.
(c) The space between the double slit and the screen is

filled with water. Explain the changes to the interference
pattern.

Given:

d  0.05102 m D  2 m
1st bright fringe  m  1 Y 1  0.13102 m

(a) Using: Non Face-to-face SLT

Ym  m D

d

  dYm  0.05102 (0.13102 )
mD (1)2

  3.25 107 m

(b) 4th dark fringe  m  3 Y 3  ?

(m  1 ) D (3  1)(3.25107 )(2)
Ym  2 2
d  0.05102

 4.55 103 m

Non Face-to-face SLT

(c) Since index of refraction,nwater > nair , λ in water is
shorter.

From equation:

y   D

d
when   y 

Distance between 2 successive fringes decreases.
The fringes are more closely spaced.

Exercise 152 Non Face-to-face SLT

(1) In Young’s two slits experiment using red light, state
the effect of the following procedure on the appearance
of the fringes :

(a) The separation of the slits is decreased.

From : y  D  y  1

dd

Separation of fringes increases.

(b) The screen is moved closer to the two slits.

Separation of fringes decreases.

(c) The source slit is moved closer to the two slits.

Separation fringes is unaffected but their brightness
increases.

(d) Blue light is used in place of red light. Non Face-to-face SLT

λblue < λred
Separation of fringes decreases.

(e) One of the two slits is covered up.

Fringes disappear. No interference. Becomes single slit diffraction.

(f) The source slit is made wider.

Fringes gradually disappear.

(g) White light is used in place of red light.

Central fringe is white. Fringes on either side are
coloured. Blue is the colour nearer to central fringe and
red is farther away.

Exercise 153 Non Face-to-face SLT

In a Young’s slits experiment, the separation between first and fifth bright fringe
is 2.5 mm when the wavelength used is 6.2×10–7 m. The distance from the slits to
the screen is 0.80 m. Calculate the separation of the 2 slits.

2.5 mm  Y 5  Y1

2.5 103  5D  D d=? 2.5 mm
Y5 Y1
dd
D = 0.8 m
2.5 103  4D

d

2.5 103  4(6.2 107 )0.8
d

d  0.8 103 m

Exercise 154 Non Face-to-face SLT

Light from a sodium lamp with a wavelength of 598.3 nm is shone normally on a soap
film. If the film has an index of refraction n = 1.40 and is suspended in the air ( n= 1.00),
find the minimum thickness for which it appears dark in reflected light ?

Given:
nsoap = 1.40
λ = 598.3 × 10-9 m

appeared dark destructive interference

Solution Non Face-to-face SLT

Step 1 : make a drawing showing the geometry of the thin film.

✓✗

12

t ? air
Soap as the thin film

air

Step 2 : Apply the rule for phase changes to each reflected rays.

Since nsoap > nair , thus ray 1 undergoes  radian phase

change but ray 2 do not undergoes phase change.

Both rays act as 2 anti-phase coherent sources.

Non Face-to-face SLT

Step 3 : solve the resulting interference equation.

2 nt  m  m  0,1,2,3, Try substitute m = 0, This
will give
For minimum thickness, m = 1
2nt  (0)

t 0

This is not considered as
minimum thickness.

t min    598.3109  2.1368  10 7 m
2 nsoap 2 (1.40)

Non Face-to-face SLT

Exercise 155

Nearly normal incident white light falls onto a glass block whose surface is coated with a thin uniform
transparent film. The refractive index of the film is 1.38 and that of glass block is 1.52. Determine the
minimum thickness of the film which could reduce reflection of light of wavelength 500 nm.

Given:
nfilm = 1.38

nblock = 1.52
λ =500 × 10-9 m

Non Face-to-face SLT

Solution

Step 1 : make a drawing showing the geometry of the thin film.

✓✓ n  nfilm air

12  Ray 1 has  radian phase change

n  nglassfilm

Air 1.0  Ray 2 has  radian phase change

t ? thin film 1.38 Both rays act as in phase

Glass block 1.52 coherent source.

Step 2 : Apply the rule for phase changes to each reflected rays.
Step 3 : solve the resulting interference equation.
Reduce reflection of light means destructive interference

2nt  (m  1) m  0,1,2,3, Non Face-to-face SLT

2 Try substitute
m = 0, This give
For minimum thickness, m = 0 minimum
thickness of
2(1.38)t min  (0  1 )500109
2 2nt  (0  1)

t min  9.0580 108 m 2

t min  90.58 nm t 

2n

Exercise 156 Non Face-to-face SLT

A thin film of thickness 4.0×10–5 cm in air is illuminated almost normally with
a white light. If the refractive index of the film is 1.50, calculate the
wavelength of high intensity in reflected light.

High intensity  constructive interference.
Both reflected rays acts as anti phase coherent sources.

✗✓ 2nt  (m  1) m  0,1,2,

12 2

Air 1.0 2(1.50)(4 105 102 )  (m  1 )
Thin film 1.50
2

  2.4m, 800nm, 480nm,

Only 480nm is in visible light spectrum.

Exercise 157 Non Face-to-face SLT

A thin film of oil (n = 1.25) is located on a smooth, wet pavement. When
viewed perpendicular to the pavement, the film appears to be
predominantly red (640 nm) and has no blue color (512 nm). How thick
is the oil film ?

Predominantly red  constructive interference for red light.
No blue color  destructive interference for blue light.

✓✓ 2nt  (m  1 )B (1)

12 2

Air 1.0 2nt  mR (2)
Oil 1.25
(1)  (2) :
t  ? Pavement (solid surface)
(m  1 )B  mR

2
m2

Subtitute into (2) :
2(1.25)(t) (2)(640109 )
t  5.12 107 m

Non Face-to-face SLT

Another alternative solution:

Condition for constructive interference is2t  m , so thicknesses that

produce constructive interference for red (640 nm) light are
2nt  m m  0,1,2,3
t  m  m(640109 )

2n 2(1.25)
t  0, 256nm, 512nm, 768nm, 

Condition for destructive interference is2t  (m  1 ) , so thicknesses

2

that produce constructive interference for blue (512 nm) light are

(m  1 ) m(640109 )

t  2  m  0,1,2,3

2n 2(1.25)

t  102nm, 307nm, 512nm, 717nm, 

Thus, thickness that can simultaneously produce constructive in red
and destructive in blue is t = 512 nm.

Exercise 158 Non Face-to-face SLT

A commonly used lens coating materials is magnesium fluoride, MgF2,
with n = 1.38. What thickness should a nonreflective coating have for
550 nm light if it is applied to a glass with n = 1.52?

Nonreflective coating  destructive interference.
Both reflected rays act as in phase coherent sources.

✓✓ 2nt  (m  1)

12 2
2(1.38)(t)  (0  1)550 109
t ? Air 1.0
MgF2 1.38 2
Glass 1.52
  9.9638 108 m

Exercise 159 Non Face-to-face SLT

A plastic film of index of refraction 1.85 is put on the surface of a car window to
increase the reflectivity and thus to keep the interior of the car cooler. The window
glass has index of refraction 1.52. (a) What is the minimum thickness is required if
light with wavelength 550 nm in air reflected from the two sides of the film is to
interfere constructively?

Both reflected rays acts as anti phase coherent sources.

✓✗ 2nt  (m  1)

12 2
2(1.85)(t)  (0  1 )550109
Air 1.0
Thin film 1.85 2

Glass 1.52   74.3 nm

Non Face-to-face SLT

(b) It is found to be difficult to manufacture and install coatings as thin
as calculated in part (a). What is the next greatest thickness for
which there will also be constructive interference?

Next greatest thickness is corresponds to m = 1

2nt  (m  1)

2
2(1.85)(t)  (1  1 )550109

2

  222.97 nm

Non Face-to-face SLT

Exercise 160

What is the thinnest soap film (excluding the case of zero thickness)

that appears black when illuminating with light with wavelength 480

nm? The index of refraction of the film is 1.33 and there is air on both

sides of the film.

black  destructive interference.

✓✗ Both reflected rays acts as anti

12 phase coherent sources.
Thinnest soap film  m = 1

Air 1.0 2nt  m
Soap 1.33
2(1.33)(t)  1(480109 )
Air 1.0
  180.45 nm

Exercise 161 Non Face-to-face SLT

Light of wavelength 580 nm is incident on a slit of width 0.30 mm.
The observing screen is 2.0 m from the slit. Find (i) the position
of the first dark fringe and (ii) the width of the central bright
fringe.

Given:   580109 m

a  0.3103 m
D  2.0m

Non Face-to-face SLT

i) distance of the 1st Dark Fringe from the Central Bright,

Yn  nD

a

Y1  (1)(580 109 )(2)
0.3103

 3.8667 103 m

ii) the width of the Central Bright, x  2Y1

 2(3.8667 103)
x  7.7334 10-3 m

Non Face-to-face SLT

Exercise 162

A sodium light of wavelength 580 nm shines through a slit
and produces a diffraction pattern on a screen 0.60 m away. The width of the
central maximum fringe on the screen is 5.0 cm. Determine
a. the width of the slit,
b. the angular width of the central maximum fringe,
c. the number of minimum that can be observed on the screen

Non Face-to-face SLT

SOLUTION

1) a. Since   580 10 9 m; D  0.60 m, w  5.0 10 2 m

1st minimum

a w Central
maximum

1st minimum

w  2 y1 and y1  D

 w  2 D  a 0.60
580 109
5.0 102 2
a a
a  1.39 105 m

b. The angular width of the central maximum fringe is given by

w  21 and 1  sin 1 

a

Non Face-to-face SLT

w  2 sin 1   w  2 sin 1  580 109 
  1.39 105
a  w  4.78

c. By applying the equation for minimum fringe,

a sin   n

For the maximum no. of order for minimum fringe,  90

   1.39 10 5 sin 90  nmax 580 10 9

nmax  23.97
 23

Therefore the number of minimum that can be observed is

23  2 = 46 fringes

Exercise 163

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 m?

SOLUTION Given,   553 109 m; a  8.00 106 m

By applying the equation for bright (maximum) fringe,

a sin    n  1 

 2

For the maximum no. of order for bright fringe,   90

Central    8.00 10 6 sin 90  nmax  0.5 553 10 9
bright
fringe nmax  13.97

 13

Therefore the number of bright that can be observed is

(13  2)+1 = 27 fringes

Non Face-to-face SLT

Exercise 164
Monochromatic light from a Helium-neon laser (λ = 632.8 nm)
is incident normally on a diffraction grating containing 6000
lines per centimeter. Find the angles at which the first order
and second order are observed.

Given: λ = 632.8×10–9 m
Diffraction grating 6000 lines per cm

Grating spacing,

d  L  1102
N 6000

d  1.6667 106 m

Non Face-to-face SLT

For the 1st order maximum, n = 1

d s in  n  n   sin n  n

d

sin  1  (1)(632.8109 ) 0.3797
1.6667 106

 1  22.31

For the 2 nd order maximum, n = 2

sin  2  (2)(632.8109 ) 0.7593
1.6667 106

 2  49.41

Exercise 165 Non Face-to-face SLT

A diffraction grating which is 2.0 cm long and has 104 lines
is illuminated normally with light of wavelength 594 nm.
How many bright lines can be observed ?

Solution

Given: 104 lines in 2 cm  2 102  2 106 m
1104
Grating spacing, d  L
N

Knowing that sin θn can not exceed unity ( 1.0 )

Highest order that we can get is given by n  d



Non Face-to-face SLT

n3 n  2 106
n2 594109

n 1 n  3.36
central max

n 1 n must be an integer number,

n  2 Highest order, n  3

n3

total number of bright lines observed = 2n + 1
= 2(3) + 1
=7

(* consists of zeroth order line, and 2 each for 1st order,
2nd order and 3rd order lines)

Non Face-to-face SLT

Exercise 166

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,
b. the angular separation between the second-order and third-order maxima.

SOLUTION Non Face-to-face SLT

35 1 N  3000 cm 1; 21  35 ; n  1

1st order max.

1st order max.

a. The diffraction angle for 1st order maximum is

21  35
1  17.5

And the slit separation, d is given by

d  L  1102 3.33 10 6 m
N 3000

Non Face-to-face SLT

Therefore the wavelength of the light is

d sin  n  n

 3.33 106 sin 1  
 3.33 106 sin 17.5  

  1.00 106 m

Non Face-to-face SLT

b. n  3 3rd order
maximum

2nd order

θ3 n  2 maximum

θ2 n0 0th order
maximum

n  2 2nd order
maximum

n  3 3rd order
maximum

By using the equation of diffraction grating for maxima,

d sin  n  n

Non Face-to-face SLT

b. For 2nd order maximum,

n2

d sin  2  2

   3.33 106 sin  2  2 1.00 106
2  36.9

For 3rd order maximum,

n3
   d sin 3  3
3.33  10 6 sin 3  3 1.00  10 6
3  64.3

Therefore the angular separation,

23  3 2
23  64.3  36.9

23  27.4

Non Face-to-face SLT

Exercise 167

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

Non Face-to-face SLT

Given : N  56 103 m 1;  2  3.1 ; n  2

(a) By applying the equation of diffraction grating for 2nd order maximum,

 thusand d  L
d sin  2  2N
sin 3.1  2 56 103 

  4.83107 m

(b) For the maximum no. of order for maximum fringe,  n  90

  sin 90  nmax 56 103 4.83 107

nmax  36.97

 36

Therefore the number of maximum can be observed is

(36  2)+1 = 73 fringes

Non Face-to-face SLT

c. The fringes become farther to each another.
Reason : since

sin   1 and d  L sin   N
dN

a larger number of lines per cm results in a larger
diffraction angle thus the distance between two consecutive
maximum fringes will increase.

LO 9.1 Planck’s Quantum Theory Non-face-to face SLT

Exercise 168

What is Planck’s Quantum Theory?

Solution

EM radiation from blackbody was emitted in discrete (separate) packets of
energy. Each packet is called a quantum of energy. This means the energy of EM
radiation is quantized. Not all values of energy are possible.

LO 9.1 Planck’s Quantum Theory Non-face-to face SLT

Exercise 169

What is the energy of a photon of yellow light of wavelength 500
nm?

Solution

(a) Using:

ℎ 6.63 × 10−34 3 × 108 = 3.798 × 10−19
= ℎ = λ =
500 × 10−9

Converting to electron–volts, we have:

3.978 × 10−19
= 1.6 × 10−19 = 2.49 eV

LO 9.1 Planck’s Quantum Theory Non-face-to face SLT

Exercise 170

Give two differences between Planck’s quantum theory and classical
theory of energy.

Solution

Classical theory Plank’s theory

The distribution of energy in Energy depends on the frequency.
blackbody depends on the
temperature.

Energy of EM radiation is emitted Energy of EM radiation is emitted in
continuously. discrete packet.

LO 9.1 Planck’s Quantum Theory Non-face-to face SLT

Exercise 171

(a) What is the energy of a photon of red light of wavelength 650 nm?
(b) What is the wavelength of a photon of energy 2.40 eV?

Solution

(a) Using: E  hf  hc 6.6631503140(309180)


E3.061019J

Converting to electron–volts, we have: 1 eV  1.61019 J

E  3.06 1019 1.91 eV
1.6 10 19

(b) Converting energy in eV to Joule (J)

E  2.40 eV  2.4 (1.61019)  3.84 1019J
E  3.841019J

Using: E  hf  hc   hc

E

6.633.1804314(3011980)

  5.18107 m

LO 9.1 Planck’s Quantum Theory Non-face-to face SLT

Exercise 172

The wavelength of a beam radiation is 400 nm. The total energy of the
photons in the beam is 0.03 J. How many photons are there in the radiation?

Solution

ℎ
= λ

0.03 = 6.63 × 10−34 3 × 108

400 × 10−9

= 6.033 × 1016 ℎ