EP025 Note #KMKK

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CHAPTER 6

GEOMETRICAL OPTICS

LEARNING OUTCOMES
At the end of this chapter, students should be able to:

6.1 Reflection at a spherical surface

a) State radius of curvature, R=2f for spherical mirror.

b) Sketch ray diagrams with a minimum of two rays to determine the characteristics of

image formed by spherical mirrors.

c) Use mirror equation, 1 = 1 + 1 = 2 for real object only.


* Sign convention for focal length, f:

(i) Positive f for concave mirror.

(ii) Negative f for convex mirror.

d) Define and use magnification, = ℎ = − respectively.
ℎ

6.2 Refraction at a plane and spherical surfaces

a) Use 1 + 2 = 2− 1 for spherical surface.


* Sign convention for radius of curvature, R:

(i) Positive R for concave mirror.

(ii) Negative R for convex mirror.

6.3 Thin lenses

a) Sketch ray diagrams with a minimum of two rays to determine the characteristics of
image formed by concave and convex lenses.

b) Use thin lens equation, 1  1  1 for real object only.
uv f

* Sign convention for focal length, f:

(i) Positive f for concave mirror.

(ii) Negative f for convex mirror.
c) Use lens maker’s equation,

1   nmaterial 1 1  1 
f nmedium r1 r2

∗ = 1

d) Define and use magnification, = ℎ = − respectively.
ℎ

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6.1 Reflection at a spherical surface

1. Figure 6.1 shows a concave mirror placed on a table. An object is held at a distance of 16.0
cm from the mirror and an image is produced as shown. What is the radius of curvature of this
mirror?

Figure 6.1

A. 8 cm
B. 16 cm
C. 32 cm
D. 64 cm

2. If an object is 6 cm in front of a concave mirror with a focal length of 4 cm, its image is at
A. 4 cm
B. 6 cm
C. 8 cm
D. 12 cm

3. Which of the following statements about convex and concave mirrors is TRUE?

A. A convex mirror can only produce real images for real objects.
B. The radius of curvature of a convex mirror is equal to its focal length.
C. A concave mirror causes a parallel beam of light incident on it to diverge.
D. A concave mirror can produce both of real and virtual images for real objects.

4. Which of the following is the best way to produce an upright, virtual and magnified image
using a concave mirror?
A. Placing an object at the center of curvature of the mirror.
B. Placing an object at a distance greater than the radius of curvature of the mirror.
C. Placing an object at a distance between the center of curvature and the focal point of
the mirror.
D. Placing an object between the focal point and the vertex of the mirror.

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5. Draw a suitable ray diagram for concave mirror for an object O placed at
i. u > 2f,
ii. u = 2f,
iii. f < u < 2f

in front of the mirror where u is the object distance and f is the focal length of the mirror.

6. An object is placed 15 cm from a
(a) concave mirror
(b) convex mirror

of radius of curvature 20 cm. Calculate the image position and magnification in each case.
Sketch the ray diagram of each case.

7. (a) A concave mirror forms an inverted image four times larger than the object. Find the
focal length of the mirror, assuming the distance between the object and the image is
0.60 m.

(b) A convex mirror forms a virtual image half the size of the object. Assuming the
distance between image and object is 20.0 cm, determine the radius of curvature of
the mirror.

8. A concave mirror has a radius of curvature 30 cm. An object of height 12 cm forms an
enlarged image 20 cm from the mirror.
(i) Calculate the object distance.
(ii) Calculate the height of the image.
(iii) State the characteristics of the image.

9. An object 4 cm high is placed 15 cm from a convex mirror of focal length 5 cm. Determine
(i) the position of the image from the mirror.
(ii) the magnification of the image.
(iii) the height of the image.

10. A dentist uses a small mirror attached to a thin rod to examine one of your teeth. When the
tooth is 1.20 cm in front of the mirror, the image it forms is 9.25 cm behind the mirror.
Determine
a) the focal length of the mirror and state the type of the mirror used,
b) the magnification of the image.

6.2 Refraction at a spherical surface
11. A goldfish is swimming inside a spherical plastic bowl of water, with an index of refraction of

1.33. If the fish is 20.0 cm from the wall of the 40.0 cm radius bowl, where does it appear to
an observer outside the bowl?

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12.

Figure 6.2
An object is placed 12 cm in front of a glass rod with hemispherical tip as shown in FIGURE
6.2. If the radius of curvature and refractive index of glass is 24 cm and 1.52 respectively,
(i) Calculate the image distance.
(ii) State TWO image characteristics.

13.

Figure 6.3
A coin is embedded in a solid glass sphere of radius 30 cm as shown in FIGURE 6.3. The
refractive index of the glass sphere is 1.5 and the coin is 20 cm from the surface.
(i) Calculate the image distance.
(ii) State the location of the image
14. FIGURE 6.4 shows an object O placed at a distance 20.0 cm from the surface P of a glass
sphere of radius 5.0 cm and refractive index of 1.63.

Figure 6.4
Determine the position of the image formed by the surface P of the glass sphere.
(Given the refractive index of air , na= 1.00)

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15. A small strip of paper is pasted on one side of a glass sphere of radius 5 cm. The paper is then
view from the opposite surface of the sphere. Determine the position of the image. (Given the
refractive index of glass =1.52 and the refractive index of air =1.00)

16. A point source of light is placed at a distance of 25.0 cm from the centre of a glass sphere of
radius 10 cm. Determine the image position of the source.
(Given the refractive index of glass =1.52 and the refractive index of air =1.00)

6.3 Thin lenses

17. Which of the following lenses is suitable for producing a real, inverted image of
magnification 1 on a screen placed at a distance of 160 cm from an object?
A. Concave lens of focal length 80 cm.
B. Concave lens of focal length 40 cm.
C. Convex lens of focal length 80 cm.
D. Convex lens of focal length 40 cm.

18. When an object is located at the focus of a concave lens, the image formed is
A. Upright
B. Inverted
C. Diminished and inverted
D. Same size as the object

19. An object is placed in front of a thin convex lens with a focal length of f. The image
produced is inverted and is half the size of the object. The object distance, u is
A. ½ f
B. 1/3 f
C. 2f
D. 3f

20. You are given a thin diverging lens. You find that a beam of parallel rays spreads out after
passing through the lens as though all the rays came from a point 20.0 cm from the center of
the lens. You want to use this lens to form an erect virtual image that is 1/3 the height of the
object. Where should the object be is placed?

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21. (a) A convex lens has a focal length of a 10 cm. Where an object is must be placed to produce
an image that is inverted and 2 times the size of the object?
(b) A diverging lens forms an image a quarter of the size of an object located 20 cm from
the lens. Find the focal length of the lens.
(c) Sketch the ray diagram of each case in part (a) and (b).

22. A meniscus convex lens has surfaces with radii of curvature 20 cm and 50 cm.
(i) Calculate the focal length.
(ii) The lens is then immersed in water with refractive index of 1.33. Calculate the new focal
length.

23.

Figure 6.5
FIGURE 6.5 shows a plano-concave lens with index refraction 1.51 and radius of curvature
18 cm. Calculate focal length of the lens.

24. (i) Write the lens maker equation and explain the sign convention used for the radii of
curvatures.
(ii)

Figure 6.6
FIGURE 6.6 shows a converging lens made of material with refractive index 2.26. One of
the surface of the lens is plane and the other has a radius of curvature of 14.5 cm.
(a) Determine the focal length of the lens.
(b) A diverging lens has a focal length f. An object is placed at position 2f from the lens.
Sketch a labeled ray diagram to determine the characteristics of the image.
25. The radii of curvature of the faces of a thin concave meniscus lens of material of refractive
index 3/2 are 20 cm and 10 cm as shown in FIGURE 6.7. What is the focal length of lens
a) in air,
b) when completely immersed in water of refractive index 4/3?

Figure 6.7

105

7.0 Physical Optics front bends or diffracts around the edges -
7.1 Huygens’s principle applied to all kinds of waves.

a) State Huygens’s principle (e.g. Figure 7.2
spherical and plane wave fronts). 7.1.b: Application of Huygens’ principle
(i) Construction of new wave front for
b) Sketch and explain the wave fronts
of light after passing through a a plane wave
single slit and obstacle using If the wave speed is v, hence in time t the
Huygens’s principle. distance travels by the wavelet is s = vt.
From Huygens’ Principle, points P1, P2, P3
7.1.a: Huygens’s principle and P4 on the wave front AB are the
Huygens’s principle states that every sources of secondary wavelets.
point on a wave front can be considered From the points, draw curves of radius s.
as a source of secondary wavelets that Then draw a straight line A’B’ which is
spread out in the forward direction at tangent to the curves at points Q1, Q2, Q3
the speed of the wave. The new wave and Q4
front is the envelope of all the secondary Hence, line A’B’ is the new wave front
wavelets - i.e. the tangent to all of them. after t second.

Figure 7.1

Diffraction of wave at a single slit Figure 7.3

Huygens’s principle can be used to explain
the diffraction of wave.
Each of the point in Figure 7.1, acts as a
secondary source of wavelets (red circular
arc)
The tangent to the wavelets from points 2,
3 and 4 is a plane wave front.
But at the edges, points 1 and 5 are the last
points that produce wavelets.
Huygens’ principle suggest that in
conforming to the curved shape of the
wavelets near the edges, the new wave

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(ii) Construction of new wave front for For light, a bright or a dark region will be
a circular wave produced in accordance to the Principle of
superposition.
Explanation as in the construction of new
wave front for a plane wave front. Principle of superposition states the
But the wave front A’B’ is a curve resultant displacement at any point is the
touching points Q1, Q2, Q3 and Q4. vector sum of the displacements due to the
The curve A’B’ is the new (circular) wave two light waves.
front after t second.
Constructive interference is a
Figure 7.4 reinforcement of amplitudes of light waves
that will produce a bright fringe
7.2 Constructive and destructive (maximum).
interferences
Destructive interference is a total
a) Define coherence cancellation of amplitudes of light waves
b) State the conditions for that will produce a dark fringe (minimum).
Monochromatic light is a light of a single
interference of light. wavelength or freqency.
c) State the conditions of constructive
7.2.b :Conditions for interference of
and destructive interference. light

Interference of light 1. Permanent interference between
Light wave is an electromagnet waves two sources of light only take place
(emw). if they are coherent sources. It
Interference is the effect of interaction means
between two or more waves which (i) the sources must have the same
overlaps or superposed at a point and at a wavelength or frequency.
particular time from the sources. (ii) the sources must have a
constant phase difference
between them.

2. The light waves that are interfering
must have the same or
approximately of amplitude to
obtain total cancellation at
minimum or to obtain a good
contrast at maximum.

3. The distance between the
coherent sources should be as
small as possible of the light
wavelength (  ).

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7.2.c : Conditions of constructive and
destructive interference
Path difference, L the difference in
distance from each source to a
particular point.

Path difference, L = |S2P  S1P| Path difference for destructive
= |x2 –x1| interference

(i) Interference of two coherent sources S1 and S2 are two coherent sources in
IN PHASE phase

Path difference for constructive
interference
S1 and S2 are two coherent sources in
phase

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Interference pattern for two coherent
sources in phase

(ii) Interference of two coherent sources Path difference for destructive
IN ANTIPHASE interference

Path difference for constructive S1 and S2 are two coherent sources in
interference antiphase
S1 and S2 are two coherent sources in
antiphase

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Interference pattern for two coherent ii) ym  (m  1 )D for dark
sources in antiphase 2

Table 2.1 shows the summary of chapter d
7.2.c.
fringes (minima),

where m = 0, ±1, ±2, ±3, … .

b) Use  y   D and explain the
d

effect of changing any of the
variables.

Methods of obtaining two coherent
sources

Division of wave front

7.3 Interference of transmitted light Figure 7.5
through double-slits
A slit S is placed at equal distance from
a) Use slits S1 and S2 as shown in figure.
Light waves from S that arrived at S1 and
i) ym  mD for bright fringes S2 are in phase.
d Therefore, both slits S1 and S2 are two new
coherent sources, e.g. in Young’s double
slit experiment

(maxima)

110

Division of amplitude S1 and S2 are produced two new sources of
coherent waves in phase because they
Figure 7.6 originate from the same wave front and
The incident wave front is divided into two their distance from S are equal.
waves by partial reflection and partial
transmission. An interference pattern consisting of bright
Both reflected waves 1 and 2 are coherent and dark fringes is formed on the screen as
and will result in interference when they shown in Figure 7.7.
superpose. e.g. Newton’s ring, air wedge
fringes and thin film interference. The bright fringes are occurred when the
7.3.a : Young’s double-slit experiment light from slits S1 and S2 superposes
constructively.

The dark fringes are occurred when the
light from slits S1 and S2 superposes
destructively.

(i) Equation for separation between
central bright fringe and mth bright
fringe

Figure 7.7
Explanation of Young’s double-slit
experiment by using Huygens’s
principle

Wave front from light source falls on a
narrow slit S and diffraction occurs.
Every point on the wave front that falls on
S acts as sources of secondary wavelets
that will produce a new wave front that
propagate to slits S1 and S2.

111

(ii) Equation for separation between Appearance of Young’s double-slit
central bright fringe and mth dark experiment
fringe
From the equation:
7.3.b : Equation for separation between
successive (consecutive) bright y  D
or dark fringes, y d

Δy: separation between two consecutive y depends on :
dark or bright fringes
the wavelength of light, 

the distance apart, d of the double slits,

distance between slits and the screen, D

Explanation for the above factors:

if  is short and thus Δy decreases for
fixed D and d. The interference fringes are
closer to each other and vice-versa.

if the distance apart d of the slits
diminished, Δy increased for fixed D and 
and vice-versa.

if D increases Δy also increases for fixed 
and vice-versa.

if a source slit S (Figure 7.7) is widened
the fringes gradually disappear. The slit S
then equivalent to large number of narrow
slits, each producing its own fringe system
at different places. The bright and dark
fringes of different systems therefore
overlap, giving rise to a uniform
illumination.

if one of the slit, S1 or S2 is covered up, the
fringes disappear.

112

if the source slit S is moved nearer the Example 1:
double slits, Δy is unaffected but their
intensity increases.

if the experiment is carried out in a
different medium, for example water, the
fringe separation Δy decreased or
increased depending on the wavelength, λ
of the medium.

if white light is used the central bright
fringe is white, and the fringes on either
side are coloured. Violet is the colour
nearer to the central fringe and red is
farther away as shown in Figure 7.8.

Figure 7.8
Table shows the range of wavelength for
colours of visible light.

113

Example 2: Interference due to reflected waves is
observed in many everyday circumstances
such as bright colours reflected from oil
film on water and soap bubble.

The reflected waves can change their
phase in two ways:
The phase changes in proportion to the
distance of the waves travel.
The phase changes as a result of the
reflection process itself.

Optical path is defined as the product
between a distance travelled by light
and the refractive index of the medium
Or

7.4 Interference of reflected light in thin 7.4.a : Phase changes due to reflection
films
A light wave travelling in a medium of
a) Identify the occurrence of phase lower refractive index (n1) when
change upon reflection. reflected from a medium’s surface of
higher refractive index (n2) undergoes a
b) Describe with the aid of a diagram  radian phase change as shown.
the interference of light in thin
films at normal incidence.

c) Use the following equations for
reflected light with no phase
difference (nonreflective coating)
i) Constructive interference:
2nt = mλ
ii) Destructive interference:
2nt = (m + ½ )λ

d) Use the following equations for
reflected light of phase difference 
rad (reflective coating)
i) Constructive interference:
2nt = (m + ½ )λ
ii) Destructive interference:
2nt = mλ
where m = 0, ±1, ±2, ±3,

e) Explain the application of thin
films (eg: solar panel, glass tint)

114

A light wave travelling in a medium of point D very close to B (BC and CD
higher refractive index (n2) when become straight line).
reflected from a medium’s surface of
lower refractive index (n1) undergoes no At B,
phase change as shown.
the reflected ray (ray 1) undergoes 
7.4.b/c : Interference from thin films for radian phase change.
reflected light with no phase
difference because the ray 1 reflected from a surface
of higher refractive index (denser
medium).

At C,

the reflected ray (ray 2) undergoes 
radian phase change.

Therefore both rays 1 and 2 are two
coherent sources in phase because the
phase difference,  is

     0

and meet at a point produces interference
pattern.

Figure shows the light waves reflected 7.4.b/d : Interference from thin films for
from the upper and lower surfaces of a thin
film (refractive index, n) on a denser reflected light of phase
medium. difference π rad

When an incident ray falls on a thin film
surface almost normal to the surface (point
B)

division of amplitude occurs,

part of ray are reflected (ray 1 ray
ABE),

part of ray are refracted and reflected (ray
2 ray ABCDF),

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Figure shows the light waves reflected 7.4.e : Application of thin films on solar
from the upper and lower surfaces of a thin panel
film (refractive index, n) in a less dense Solar cells is a devices that generate
medium. electricity when exposed to sunlight are
When an incident ray falls on a thin film often coated with transparent, thin film of
surface almost normal to the surface (point silicon monoxide (SiO, n=1.45) to
B) minimize reflective losses from the
division of amplitude occurs, surface. Suppose that a silicon solar cell
part of ray are reflected (ray 1 ray (n=3.5) is coated with a thin film of silicon
ABE), monoxide for the purpose.
part of ray are refracted and reflected (ray
2 ray ABCDF), Example 3:
point D very close to B (BC and CD
become straight line).
At B,
the reflected ray (ray 1) undergoes 
radian phase change.
because the ray 1 reflected from a surface
of higher refractive index (denser
medium).
At C,
the reflected ray (ray 2) undergoes no
phase change.
Therefore both rays 1 and 2 are two
coherent sources antiphase because the
phase difference,  is

     0   rad

and meet at a point produces interference
pattern.

116

c) Use:

i) yn  nD for dark fringes
a

(minima)

ii) yn  (n  1 )D for bright fringes
2

a

(maxima),

where n = ±1, ±2, ±3, ...

7.5.a : Diffraction of light

Diffraction of light is defined as the
bending of light waves as they travel
around obstacles or pass through an
aperture or slit comparable to the
wavelength of the light waves.
Figures 7.9 show the bending of plane
wavefront.

Example 4:

Figure 7.9

7.5.b/c : Diffraction by a single silt

Figure 7.10 shows an apparatus setup of
diffraction by a single slit.

7.5 Diffraction by a single slit Figure 7.10
a) Define diffraction.
b) Explain with the aid of a diagram
the diffraction of a single slit.

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Explanation of single slit diffraction In general, for minima (dark fringes)
experiment
If the distance of single slit to the screen is
Wavefront from light source falls on a D, and D>>a then:
narrow slit S and diffraction occurs.
sin n  tan n  yn
Every point on the wavefront that falls on D
S acts as sources of secondary wavelets
and superposed each another to form an Therefore the distance of nth minimum
interference pattern on the screen as shown from central maximum is:
in Figure 7.10.
(ii) : Equation for separation between
The central fringe is bright (maximum) central maximum (bright) and nth
and widen compare to other bright fringes.
The central fringe has the highest maximum (bright) fringes
intensity compare to the other bright
fringes.

The intensity of bright fringes reduce as
the distance increase from the central
bright fringe.

Other rays with angle θ2 and θ1 will
produce minimum and maximum on both
sides of the central maximum.

(i) : Equation for separation between
central maximum (bright) and nth
minimum (dark) fringes

Consider two narrow strips as shown in Consider three narrow strips as shown in
figure above, for the two strips figure above, the first two strips (pair)
superposed destructively thus both strip superposed destructively at which the path
of light must in antiphase to each another difference is ½ and leave the third strip.
which is equivalence to a path difference The 3rd strip produces the maximum
of ½. (bright) fringe at R.

If the 1st minimum (1st order minimum) If the 1st maximum (1st order maximum)
is at P, hence : is at R, hence :

In general, for maxima (bright fringes)
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where The angular width of central maximum,
If the distance of single slit to the screen is w is given by
D, and D>>a then:
The width of central maximum, w is
sin n  tan n  yn given by
D

Therefore the distance of nth maximum
from central maximum is:

Equation for central maximum (bright) To calculate the maximum number of
fringe orders observed, take the diffraction
angle, = 90.

From both equations for minima and
maxima, we obtain

By using this two relations, the changes of
single slit diffraction pattern can be
explained.

Figure 7.11 Example 5:

Figure 7.11 shows five sources of
Huygen’s wavelets and the screen is to be
so far from the slit (D>>a) thus the rays
from each source are nearly parallel.
All the wavelets from each source travel
the same distance to the point Q (Figure
2.44) and arriving there in phase.
Therefore, the constructive interference
is occurred at the central of the single
slit diffraction pattern.

119

7.6 Diffraction grating

a) Explain with the aid of a diagram

the formation of diffraction.

b) Apply d sin θ = nλ where = 1


(eg: transparent compact disk,

muslin cloth, etc)

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

Diffraction grating can be made by ruling
very fine parallel lines on glass or metal
by a very precise machine.

The untouched spaces between the lines
serve as the slits as shown in Figure 2.46.

Example 6:

If there N lines per unit length, then slit
separation, d is given by:

e.g. if a diffraction grating has 5000 lines
per cm, then

The light that passes through the slits are
coherent .
The Interference pattern is narrower and
sharper than double-slits.
There are two type of diffraction grating
which are

(i) transmission grating (usual
diffraction grating)

(ii) reflection grating e.g. CD and
DVD

120

Diffraction grating is used in spectrometer 7.6.b : Equation of diffraction grating
to determine the wavelength of light and
to study spectra. Figure 2.48 illustrates light travels to a
distant viewing screen from five slits of
Explanation of diffraction by using the grating.
Huygens’s principle for diffraction
grating

Figure 7.12 shows an incident lights fall
on the transmission diffraction grating.

Figure 7.13

Figure 7.12 The maximum (bright) fringes are
Using Huygens’ principle, each maximum sometimes called the principal maxima or
is located by taking the tangent of the principal fringes since they are placed
wavelets from the slits. where the light intensity is a maximum.

If the wavelets from each of the slits are Since the screen is far so that the rays
drawn and a tangent AB is drawn, a plane nearly parallel while the light travels
wave front parallel to the diffraction toward the screen as shown in Figure 7.13.
grating is obtained. This represents the In reaching the place on the screen while
zeroth-order maximum (n = 0). the 1st order maximum is located, light
from one slit travels a distance of one
If the wavelets are grouped such that the wavelength farther than light from
first wavelet from one slit is combined adjacent slit.
with the second wavelet from the next slit,
the third wavelet from the third slit and so Therefore the path difference for
on, the tangent CD will represent the first-
order maximum (n =1). maximum fringe (constructive

For the second-order maximum, the interference) is given by
wavelets are grouped are such that the
second wavelet of one slit is combined
with the fourth wavelet of the next slit, the
sixth wavelet from the third slit and so on.
(tangent EF)

Similarly, the third-, fourth-,…. order
maximum may be obtained

121

The maximum fringes produce by a
grating are much narrower and sharper
than those from a double-slit as the
intensity graph in Figures 7.14

Figure 7.15

To calculate the maximum number of
orders for bright fringes observed, take
the diffraction angle, = 90. Therefore

Figure 7.14
Figures 7.14 shows the diffraction grating
pattern.

From the equation for maxima, we obtain

If the white light is falls on the grating, a By using this two relations, the changes of
rainbow colours would be observed to diffraction grating pattern can be
either side of the central fringe on the explained.
screen which is white as shown in Figure
7.15. This because the white light Example 7:
contains wavelengths between violet and
red.

122

Example 8:
123

8.0 Physical Optics

Learning Outcomes:

7.1 Huygens’s principle

c) State Huygens’s principle (e.g. spherical and plane wave fronts).
d) Sketch and explain the wave fronts of light after passing through a single slit and

obstacle using Huygens’s principle.

7.7 Constructive and destructive interferences

d) Define coherence
e) State the conditions for interference of light.
f) State the conditions of constructive and destructive interference.

7.8 Interference of transmitted light through double-slits

c) Use

i) ym  mD for bright fringes (maxima)
d

ii) ym  (m  1 )D for dark fringes (minima),
2

d
where m = 0, ±1, ±2, ±3, … .

d) Use  y   D and explain the effect of changing any of the variables.
d

7.9 Interference of reflected light in thin films

f) Identify the occurrence of phase change upon reflection.

g) Describe with the aid of a diagram the interference of light in thin films at normal

incidence.

h) Use the following equations for reflected light with no phase difference

(nonreflective coating)
i) Constructive interference: 2nt = mλ
iii) Destructive interference: 2nt = (m + ½ )λ

i) Use the following equations for reflected light of phase difference  rad

(reflective coating)
iii) Constructive interference: 2nt = (m + ½ )λ
iv) Destructive interference: 2nt = mλ

where m = 0, ±1, ±2, ±3,

j) Explain the application of thin films (eg: solar panel, glass tint)

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7.10 Diffraction by a single slit

d) Define diffraction.

e) Explain with the aid of a diagram the diffraction of a single slit.

f) Use:

i) yn  nD for dark fringes (minima)
a

ii) yn  (n  1 )D for bright fringes (maxima),
2

a

where n = ±1, ±2, ±3, ...

7.11 Diffraction grating

c) Explain with the aid of a diagram the formation of diffraction.

d) Apply d sin θ = nλ where = 1 (eg: transparent compact disk, muslin cloth, etc)


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7.1 Huygens’s principle
1. One evidence that shows that light is a wave phenomenon is that

A. Light can be reflected
B. Light can be refracted
C. Light can undergo diffraction and interference.
D. Light can transport energy

2. Which of the following methods will result in a new wavefront?
A joining all the tangents to the wavelets
B joining all the normal to the wavelets
C joining the lines perpendicular to the wavelets
D joining the circles which have centres on the wavefront

3. Which of the following statements shows the relationship between the wavefront
and the direction of the wave?
A the wavefront is parallel to the direction of the wave
B the wavefront is perpendicular to the direction of the wave
C The wavefront does not depend on the direction of the wave
D the wavefront is circular when the waves travel in one direction

4. Huygens’ principle can be used to construct wavefronts in the following cases
except
A resolution
B reflection
C refraction
D diffraction

5. Which of the following is a false statement?

A All points on a given wave front have the same phase.
B Rays are always perpendicular to wave fronts.
C All wave fronts have the same amplitude.
D The spacing between adjacent wave fronts is one-half wavelength.

6. (a) What is meant by wave front?
(b) State the Huygens’ Principle.

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(c) Sketch the diffraction patterns produced when the wave fronts pass through
the obstacle in each given diagram.
i.

ii.

iii.

7.2 Constructive and destructive interferences
7. Which of the following statement is TRUE concerning two monochromatic light

waves that are coherent?
A. They have same phase
B. They have constant phase difference
C. They have almost the same amplitude
D. They have constructive disturbances
8. Constructive interference occurs at a point because
A two coherent waves have the same intensity
B two coherent waves arrive at that point at the same time
C two coherent waves have the same maximum amplitudes at that point
D the path difference between two coherent waves is a multiple of their

wavelengths

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9. (a) Define coherence.
(b) State the conditions for superposition of two light sources to produce an
interference pattern.
(c) State the conditions of constructive and destructive interference.

7.3 Interference of transmitted light through double-slits

10. The FIGURE 18.1 shows the arrangement of Young’s double-slit experiment. S
is the monochromatic light source of wavelength, λ. The centre of the interference
pattern which is produced on the screen is located at O with OS being
perpendicular to the two slits S1 and S2. P is the location of the second dark
fringe.

FIGURE 23.1

Which of the following is true about the phase difference between the waves from

S1 and S2 and the optical path difference between S1P and S2P

Phase difference Optical path difference
A. 0
1 1
B. 0 2
C. 3
21
D. 3 2

1 1
2

21
2

11. Light of wavelength 575 nm falls on a double-slit and the third order bright fringe
is seen at an angle of 6.5°. What is the separation between the double slits?

A 5.0 μm
B 10 μm
C 15 μm
D 20 μm

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12. The figure below shows how the apparatus for a double-slit experiment are

arranged. The distance between the slits is d, and the distance between the slits

and the screen is D. An interference pattern is produced on the screen when a
light of wavelength, λ is incident on the slits.

FIGURE 7.1
(i) If λ = 5.4 x 10-7 m, D = 1.8 m and the 9th bright fringe is formed at a

distance 4.0 cm from the center. Calculate the slits separation.
(ii) A laser beam (λ = 632.8 nm) is incident on two slits 0.200 mm apart. How

far apart are the bright interference fringes on a screen 5.00 m away from

the double slits?

13.

Monochromatic S1
S
Source *
S2

Slits Screen

The above arrangement is used to produce Young’s fringes on the screen from a
monochromatic source.
(i) Slit S should be narrow to produce Young’s fringes. Explain why.
(ii) Why do slits S1 and S2 act as coherent sources?

7.4 Interference of reflected light in thin films

14. When the glass windows of a house are pasted with thin film, we cannot see the
people in the house during the day. When the lights are switched on the night, we
can see the people in the house. This is because
A we can see the people when the lights are switched on
B the reflected lights undergo destructive interference
C we can see the people by the refracted light through the glass only
D the light cannot pass through the thin film pasted on the glass windows

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15. When a beam of monochromatic light of wavelength 500 nm is incident on a soap
bubble, the soap bubble appears black in color. The refractive index of the soap
bubble is 1.40. What is the thickness of the soap bubble?
A 54 nm
B 89 nm
C 102 nm
D 179 nm

16. A thin plastic sheet of refractive index 1.60 is pasted on a glass plane to reduce
infrared losses in a farming laboratory. The wavelength of the infrared radiation is
700 nm. Determine the minimum thickness of the plastic sheet necessary to
produce constructive interference in the reflected infrared radiation
A 55 nm
B 109 nm
C 218 nm
D 318 nm

17. Light from a sodium lamp with a wavelength of 589.3 nm is shone normally on a
soap film. If the soap film appeared dark when viewed from the incident side.
What is the minimum thickness of the soap film? (Assume the refractive index of
the soap is 1.4)

18. A glass lens is coated with a thin film of a transparent composite in order to make
the lens non-reflecting. What will the minimum thickness of the film be if the
refractive index of glass is 1.5, refractive index of the composite material is 2.82
and the wavelength of the light used is 610 nm?

7.5 Diffraction by a single slit

19. In a single slit diffraction experiment, if the width of the slit increases, what
happens to the width of the central maximum on a screen?
A It increases.
B It decreases.
C It remains the same.
D There is not enough information to determine.

20. A single slit, which is 0.050 mm wide, is illuminated by light of 550 nm
wavelength. What is the angular separation between the first two minima on
either side of the central maximum?
A 0.36°
B 0.47°
C 0.54°
D 0.63°

21. A beam of green light is diffracted by a slit of width 0.550 mm. The diffraction
pattern forms on a wall 2.06 m beyond the slit. The distance between the positions
of zero intensity on both sides of the central bright fringe is 4.10 mm. Calculate
the wavelength of the laser light.

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22. State the similarities and differences of double-slit interference and single slit
diffraction patterns.

23. 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

24. A beam of monochromatic light is incident normally on a single slit of width 0.2
mm. the third dark fringe is located 1.40 cm from the central bright fringe on a
screen 120 cm from the slit.
(a) Calculate the wavelength of the light used
(b) Explain what happens when the width of the single slit is increased.

7.6 Diffraction grating

25. Monochromatic light is incidence normally on a diffraction grating with 500 lines
per mm. The second-order line is seen at an angle of 30° with the normal. What is
the wavelength of the light?
A 0.25 µm
B 0.50 µm
C 0.58 µm
D 0.65 µm

26. A diffraction grating has 5000 lines per cm. The angle between the central
maximum and the fourth order maximum is 47.2°. What is the wavelength of the
light?

A 138 nm
B 183 nm
C 367 nm
D 637 nm

27. A beam of monochromatic light of wavelength 600 nm is incident normally on a
diffraction grating which has 500 lines per mm. How many spectral lines can be
observed on the screen?
A2
B3
C5
D7

28. A beam of monochromatic light of wavelength 589 nm is incident on a diffraction
grating. The second order maximum is formed at an angle of 15o from the central
maximum.
(a) Calculate the number of lines per mm on the diffraction grating
(b) How many spectral lines could be observed?

29. A parallel beam of white light of wavelength 450 nm to 720 nm is incident
normally on a diffraction grating which has 3.0 x 105 lines per metre.
(a) Calculate the angular width of the second order spectrum
(b) What is the highest order of the diffraction spectrum that can be observed?

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CHAPTER 8

QUANTIZATION OF LIGHT

LEARNING OUTCOMES
8.1 PLANCK’S QUANTUM THEORY
a) State Planck’s quantum theory.
b) Distinguish between Planck’s quantum theory and classical theory of energy.

c) Use Einstein’s equation for photon energy , E = hf = ℎ



8.2 PHOTOELECTRIC EFFECT

a) Define photo electric effect.

b) Explain the phenomenon of photoelectric effect by sketching diagram of the photoelectric effect
experimental set-up.

c) Define threshold frequency, work function and stopping potential.

d) Analyze by using graph and equations the observations of photoelectric effect experiment in terms
of the dependence of:

i. kinetic energy of photoelectron on the frequency of light;

1 mvmax2  eVs  hf  hf0 photoelectric current on intensity of incident
2

light;

ii. work function and threshold frequency on the types of metal surface;

W0  hf0 .

e) Use Einstein’s photoelectric equation , Kmax  eVs  hf W0

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8.1 PLANCK’S QUANTUM THEORY
1. A photon is

A. a very narrow beam of light.
B. a beam of sunlight
C. a portion (quantum) of electromagnetic energy.
D. the name given to a peak value in the blackbody spectrum
2. The probability for stimulated emission depends
A. only on the frequency of radiation.
B. only on the structure of an atom.
C. only on the energy density of radiation.
D. on the structure of an atom and on the energy density of radiation.

3. What is the energy of a photon of red light of wavelength 650 nm?
4. A radio station broadcasts at 88.3 MHz. If the power is 50 Megawatts, how many photons per

second are emitted from the antenna?
5. State ONE characteristic of electromagnetic wave energy that contradicts the Planck’s

quantum theory and classical theory.
6. Show that the photons in a 1240 nm infrared light beam have energies of 1.00 eV.
7. Compute the energy of a photon of blue light of wavelength 450nm.
8. To break a chemical bond in the molecules of human skin and thus cause sunburn, a photon

energy of about 3.50 eV is required. To what wavelength does this correspond ?
9. What is the energy in electron volts of a photon of green light of wavelength 546nm?
10. Find an expression for the energy of a photon in electron volts when the wavelength of the

photon is given in nanometres.
11. Calculate the energy, in electron volts, of a photon whose frequency is

i. 620 THz,
ii. 3.10 GHz,
iii. 46.0 MHz.

iv. Using information on question (i) , (ii) and (iii) Determine the corresponding
wavelengths for those photons and state the classification of each on the
electromagnetic spectrum.

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8.2 PHOTOELECTRIC EFFECT

12. In the photoelectric effect,
A. the energies of electrons liberated by light depend on the frequency of the
light,
B. the energies of electrons liberated by light depend on the source of the light,
C. the energies of electrons liberated by light depend on the intensity of the light.

13. The work function for Tungsten metal is 4.25 eV. What is the threshold wavelength, λo for
Tungsten?

14. Ultraviolet waves with frequency from 8.00 × 1015 Hz to 1.00 ×1017 Hz are incident on a zinc
surface. The work function of zinc is 4.33 eV. Calculate the maximum kinetic energy of
electrons ejected from the surface.

15. (a) The Einstein’s Equation is given by where Wo is the work function of the metal.
Explain the physical process this equation represents.

(b) The longest wavelength of electromagnetic radiation that can emit electrons from the
surface of Sodium is 500 nm. What is the work function of Sodium?

16. (a) Explain briefly the phenomenon of photoelectric effect
(b) Describe and sketch diagram of the photoelectric effect experimental set-up.
(c) Define work function, threshold frequency and stopping voltage.
(d) Einstein’s equation for photoelectric effect is Kmax = hf – W. Explain the meaning of
each term in the equation.
(e) Sketch the graph of the maximum kinetic energy against frequency of radiation for 3
different metal surfaces.

17. (a) A clean surface of potassium in a vacuum is irradiated with light of wavelength
5.50x10-7m and electrons are found just to emerge, but when light of wavelength
5.00x10-7m is incident, electrons emerge each with energy 3.65x10-20J . Calculate the
value for Planck’s constant h.

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(b) Molybdenum has a work function of 4.20 eV.
i. Find the cutoff/threshold wavelength and cutoff/threshold frequency for the
photoelectric effect.

ii. What is the stopping potential if the incident light has a wavelength of
180nm?

18. (a) Electrons are ejected from a metallic surface with speeds ranging up to 4.60×105m/s
when light with a wavelength of 625 nm is used.
i. What is the work function of the surface?
ii. What is the cutoff frequency for this surface?

(b) Lithium, beryllium, and mercury have work functions of 2.30 eV, 3.90 eV and 4.50
eV, respectively. Light with a wavelength of 400 nm is incident on each of these
metals. Determine :
i. which metals exhibit the photoelectric effect and
ii. the maximum kinetic energy for the photoelectrons in each case.

19. In a photoelectric effect experiment, ultraviolet light of wavelength 254nm with intensity of
210W m-2 shines onto a metal with surfac of 12 mm2. The photoelectric current recorded is
4.89 × 10-10 A.
(a) Calculate the number of photons incident onto the metal each second.
(b) How many photoelectrons are emitted per second? Assuming all the photoelectrons
emitted are received by the collector electrode.
(c) When the experiment is repeated with ultraviolet light of wavelength 313 nm, no
electrons are emitted. Explain why no photoelectrons are emitted and support your
reasoning with calculations.

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CHAPTER 9
WAVE PROPERTIES OF PARTICLE
9.1 de Broglie wavelength
At the end of this chapter, students should be able to:
a) State wave-particle duality.

b) Use de Broglie wavelength,   h .
p

9.2 Electron Diffraction
At the end of this chapter, students should be able to:

a) Describe the observations of electron diffraction in Davisson-Germer experiment
b) Explain the wave behavior of electron in an electron microscope
c) State the advantage s of electron microscope compare to optical microscpe

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