Module NF2FSLT SP025

Given : R  105 ; C  13.2106 F; L  85103 H;
f  125 Hz

(a) The capacitive reactance is

XC  1  1

2fC 2 12513.2106
 XC

X C  96.4575 

The inductive reactance is

 X L  2fL

 2 125 85103

X L  66.7588  303

Impedance of the circuit is

Z  R2  X L  X C 2
 1052  66.7588  96.45752

Z  109.1193 

power factor of the circuit is

cos  R cos  105

Z 109.1193

cos  0.9622

(b) Increase

cos  R As R increase, cos  increase.

304

TOPIC 7 :
GEOMETRICAL
OPTICS

SP025 1 hours

7.1 Reflection at a spherical surface Non Face-to-face SLT

LO 7.1 (c), (d) : Use mirror equation, = + for real object only and magnification, = = − respectively


Exercise 126

Assume that a certain spherical mirror has a focal length of +
10.0 cm. Locate & describe the image for object distances of

(a) 25.0 cm
(b) 5.0 cm in front if the mirror

Solution Focal length, f is + ve  concave mirror
(a) Given : u = + 25.0 cm
f = + 10 cm

from : 1  1  1
f uv

1 1 1 Non Face-to-face SLT
25 cm v 10 cm

1 3
v 50

Magnification, v  50  16.7 cm
3

m   v  16.7 cm  0.668
u 25 cm

m < 1.0 and negative  image is smaller than the object
(Diminished), Inverted

+ v  Image is Real

Test your understanding using question (b) Non Face-to-face SLT
Suggested Answer for (b) :
v = – 0.1 m ; m = 2.0
Image is virtual, located behind the mirror,
upright, enlarged

Exercise 127
A woman who is 1.5 m tall is located 3.0 m
from an anti shoplifting mirror. The focal
length of the mirror is – 0.25 m. Find :
(a) the position of her image
(b) the magnification
(c) the height of the image.

Given : ho = 1.5 m , u = 3.0 m ; f = –0.25 m

(a) from : 1  1  1 Put (–) sign for f
uv f because convex
1 1 1 mirror

 3.0 v  0.25

1   13  v   0.231 m
v3

(b) Magnification, m   v   (0.231)  0.077
u 3.0

= ℎ → ℎ = ℎ  0.077 (1.5)
ℎ

hi  0.12 m

Non Face-to-face SLT

Exercise 128

A pill bottle 3.0 cm tall is placed 12.0 cm in front of a
mirror. A 9.0 cm tall upright image is formed.
(a) The mirror is convex or concave? Why ?
(b) What is its radius of curvature ?

Solution ho = 3.0 cm ,
Given : hi = 9.0 cm
u = 12.0 cm,
Upright  virtual image

Image is enlarged & upright  concave mirror Non Face-to-face SLT
Reason:
Only concave mirror formed enlarged, upright
image when u < f
convex mirror  image always reduced in size.

(b) Magnification :

= − = ℎ
ℎ

9
− 12 = 3

= −36

from : 1  1  1 Non Face-to-face SLT
uv f
(+) sign for f
1 1 1 because concave
12  36 f mirror

(–) sign for v 1 2
f 36
because image is
upright  virtual

f  18cm

from : f R R2f
2
 2(18)
R  36 cm

LO 7.1 Reflection at a spherical surface Non-face-to face SLT

Exercise 129

A candle with flame 1.5 cm tall is placed 5 cm from the front of a concave
mirror. A virtual image is produced that is 10 cm from the vertex of the mirror.

(a) Find the focal length & radius of curvature of the mirror.
(b) How tall is the image of the flame ?

Solution

Solution

a) 1  1  1  1  1
f u v 5 (10)
f  10cm

M   v  hi
b) u h

M   (10)  hi
(5) 1.5

hi  3cm

LO 7.1 Reflection at a spherical surface Non-face-to face SLT

Exercise 130

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. Find

(a) focal length of the mirror,
(b) the magnification of the image.

Solution b) M   v   (9.25)
u 1.2
a) 1  1  1
f uv M  7.71
1 1  1
f 1.2 (9.25)
f  1.38cm

LO 7.1 Reflection at a spherical surface Non-face-to face SLT

Exercise 131

An upright image is formed 20.5 cm from the real object by using the spherical mirror. The image’s height is
one fourth of object’s height.
(a) Where should the mirror be placed relative to the object?
(b) Calculate the radius of curvature of the mirror and describe the type of mirror required.

Solution hi  1 h b) u  v  20.5cm
4 v  20.5 16.4  4.1cm
a) u  v  20.5cm
Virtual image formed, v = -4.1cm
v  20.5  u
2 11
M   v  hi R uv
u h 2 1  1
R 1.64 (4.1)
 (20.5  u) 1 h R  10.9cm
4
M   
u h

u  16.4cm

Mirror should be placed 16.4cm relative to object

LO 7.1 Reflection at a spherical surface Non-face-to face SLT

Exercise 132

a. A concave mirror forms an inverted image four times larger than the object. Calculate the focal length of
the mirror, assuming the distance between object and image is 0.600 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.

Solution 1 11 b) M   v
f uv u
a) M   v 1 1  1
u f 0.2 (0.8) 1   (20  u)
f  0.16m 2u
 4   (0.6  u) u  40cm
u
2 11
u  0.2m R uv
2 1  1
u  0.6  v R 40 (20)
0.2  0.6  v R  26.67cm
v  0.8m

LO 7.1 Reflection at a spherical surface Non-face-to face SLT

Exercise 133

a. A 1.74 m tall shopper in a department store is 5.19 m from a security mirror. The shopper notices that
his image in the mirror appears to be only 16.3 cm tall.
i. Is the shopper’s image upright or inverted? Explain.
ii. Determine the radius of curvature of the mirror.

b. A concave mirror of a focal length 36 cm produces an image whose distance from the mirror is one
third of the object distance. Calculate the object and image distances.

Solution b) u  3v
vu
a) i. upright because security mirror is convex mirror 3

a) ii. M   v  hi 2 11 1 11
u h R uv f uv
2 1  1 1 13
M   v 16.310  2 R 5.19 (0.49) 36 u u
5.19 1.74 R  1.08m u  144cm

v  0.49cm

v  u  144  48cm
33

7.2 Refraction at a spherical surface Non Face-to-face SLT

LO 7.2 (a) : Use 1 + 2 = 2− 1 for spherical surface



Exercise 134

One end of a long glass rod ( n = 1.50 ) is formed into a convex surface
of radius 6.0 cm. An object is positioned in air along the axis of the
rod. Find the image position corresponding to object distances of
(a) 20.0 cm, (b) 3.0 cm from the end of the rod.

object R = 6 cm
glass rod

Solution n1 (1.0) front Non Face-to-face SLT

object R = 6 cm
n2 (1.5) back
u = 20 cm
glass rod

Surface curves into less dense medium

u = + 20 cm ( front of the surface) ; n1 = 1.0 ; n2 = 1.5
R = + 6 cm ( surface curves into air -- less dense

medium ) −

From : + =

1  1.5  1.5 1
20 v 6

1.5  4
v 120

v  120(1.5) Non Face-to-face SLT
4

v  45cm ( Real Image )

(b) u = 3 cm (real object) −

From + =

n1 (front) R = 6 cm 1  1.5  1.5 1
n2 (back) 3 v 6
object
u = 3 cm glass rod 1.5  1.5
v6
R = + 6 cm ( convex surface ) v  6 cm

( Virtual Image )

Exercise 135 Non Face-to-face SLT

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 ?

Solution

Fish is the object. Ray travel from water into air.

Refracted ray

n2 (back) Incident ray

On1 (front)

Observer IC

R = 40 cm

sees the image of the fish

u = + 20 cm ( front of surface ) Non Face-to-face SLT

n1 = 1.33 , n2 = 1
R = −40 cm (surface is concave & in going ray from denser

to less dense medium)

From + = −


1.33  1  11.33
20 v 40

1  0.33  1.33  2.33
v 40 20 40

v  17.1cm ( Virtual Image)

Image is 17.1 cm on the same side as the object (inside the bowl)

LO 7.2 Refraction at a spherical surface Non-face-to face SLT

Exercise 136

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. Find the
position of the image.(Given refractive index of glass = 1.52 and refractive
index of air = 1.00)

Solution

n1  n2  n2  n1
uv r

1.52  1.00  1.00 1.52
10 v 5

v  20.8cm

LO 7.2 Refraction at a spherical surface Non-face-to face SLT

Exercise 137

In a second condition, the same paper is placed vertically inside another glass sphere
of radius 20.0 cm. If the refractive index of glass is 1.5 and the paper is at a distance
of 9.0 cm from the surface,

i. calculate the image distance if the paper is being
observed from the nearest surface, and

ii. state whether the image is real or virtual.

Solution

i. n1  n2  n2  n1
uv r

1.5  1.00  1.00 1.5
9v  20

v  7.1cm

ii. Virtual image formed because the image distance is -7.1cm

7.3 Thin lenses Non Face-to-face SLT

LO 7.3 (a),(b), (d),(e):

Sketch ray diagram with a minimum of two rays to determine the characteristics of image formed by concave and convex

lenses.

Use thin lens equation, 1 = 1 + 1 for real object only, lens maker’s equation and magnification, = = −


Exercise 138

A biconvex lens is made of glass whose index of refraction is 1.6 .
The lens has a radius of curvature of 30 cm for one surface and 40 cm
for the other.
(a) What is the focal length of the lens ?
(b) If the lens is immersed in water, is there a change in the focal
length of the lens ?

If so, which way ?
(c) Calculate the focal length of this lens as used under water.

Solution Non Face-to-face SLT

Given : n1 = 1.0 ( in air )
n2 = 1.6 ( lens is made of glass )
R1 = + 30 cm (convex surface)
R2 =  40 cm (concave surface)

(a) Using Lens makers’ Equation :

1  ( n2   1  1 
f n1 1) R1 R2 



 (1.6 1)[ 1  1 ]
1.0 30cm (40cm)

Non Face-to-face SLT

1  (0.6) 7
f 120

 0.035
f 1

0.035
 28.57 cm

(b) When immersed into water, focal length, f change

reason :
f depends on index of refraction of surrounding

medium, n1
when n1 greater, focal length, f become greater.

Non Face-to-face SLT

(c) Using Lens makers’ Equation :

1  ( n2  1)  1  1
f n1  R1 
 R2 

 ( 1.6 1)[ 1  1 ]
1.33  30cm  40cm

 (0.203) 7
120

 0.01184

f  1  84.46 cm
0.01184

Non Face-to-face SLT

Exercise 139

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.

(a) Where should the object be placed ?

(b) Draw a principal ray diagram.

Solution Non Face-to-face SLT

(a) f = – 20 cm = – 0.2 m
Lateral magnification,

m1
3

positive because the image is upright(virtual)

From : m   v
u

1v
3u

v   u (1)
3
: 11 1
from Thin Lens Equation uv f

Non Face-to-face SLT

1  1  1 (2)
u v  0.2

Substitute (1) into (2):

1  1   1
u [ u] 0.2

3

v is given (–) 13 1
u u 0.2
sign because
image is erect - 2 1
virtual u 0.2

u  0.4 m

Non Face-to-face SLT

Substitute u = 0.4 m into (1) we can get image distance

v   0.4  0.133 m
3

(b) Ray diagram :

u = 0.4
m

O FI F

Exercise 140 Non Face-to-face SLT

A person of height 1.75 m is standing 2.50 m in from
of a camera. The camera uses a thin biconvex lens
of radii of curvature 7.69 mm. The lens made from
the crown glass of refractive index 1.52.

(a) Calculate the focal length of the lens.

(b) Sketch a labeled ray diagram to show the
formation of the image.

(c) Determine the position of the image and its
height.

(d) State the characteristics of the image.

Solution Non Face-to-face SLT

ho  1.75 m; u  2.50 m; n2  1.52; n1  1.00;
R1  7.69 103 m (Convex surface)
R2  7.69 103 m(Concave surface)

(a) By applying the lens maker’s equation,

1  ( n2  1)  1  1 
f n1  R1 R2 
 

1  (1.52  1)[ 1  1 ]
f 1.00 7.69 103 7.69 103

f  7.39 103 m (Convex lens)

Non Face-to-face SLT

(b) The ray diagram for the case is

O2F F I

Front F 2F

back

(c) The position of the image formed is Non Face-to-face SLT

1 11 1  1  1
f uv  7.39 103 2.50 v
v  7.41103 m

(at the back of the lens)

By using the linear magnification equation, thus

= ℎ = − ℎ = − 7.41 10−3
ℎ0 1.75 2.50

hi = −5.19x10−3 m
(-ve means the image is inverted)

(d) The characteristics of the image are
 Real, inverted, diminished, formed at the back of the lens

Exercise 141 Non Face-to-face SLT

A thin plano-convex lens is made of glass of refractive index
1.66. When an object is set up 10 cm from the lens, a virtual
image ten times its size is formed. Determine

(a) the focal length of the lens,

(b) the radius of curvature of the convex surface.

Solution
n 1.66; u 10 cm; m  10 (erect/virtual)

(a) By applying the linear magnification equation for thin lens, thus

m   v  10 v  10u
u

By using the thin lens formula, thus 1  1  1
f uv

Non Face-to-face SLT

Virtual image

1  1  1
f 10
1010

f  11.1cm

(b) Since the thin lens is plano-convex and let r2  

Therefore 1  ( n2  1 1
f n1  R1
 1)   R2 


1  (1.66   1  1
11.1 1.00 1) R1 
 


R1  7.33 cm (Convex surface)

Exercise 142 Non Face-to-face SLT

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.
What is the focal length of lens

(a) in air,

(b) when completely immersed in water of refractive index
4/3?

Solution

R1  20 cm R2  10 cm

(a) By applying the lens maker’s equation,

1  ( n2   1  1
f n1 1) R1 
R2 


Given n2  3  1.00 Non Face-to-face SLT
2 ; n1

1   3 1  1  1 
f  2  20 10 

f  40 cm

(b) Givenn1  4
3

By using the general lens maker’s equation, therefore

1  ( n2   1  1
f n1 1) R1 
R2 


1    3    1  1 
f   2  1  20 10 

4
3

f  160 cm

LO 7.3 Thin lenses Non-face-to face SLT

Exercise 143

a. A glass of refractive index 1.50 plano-concave lens has a focal length of 21.5
cm is immersed in the air. Calculate the radius of the concave surface.

b. A rod of length 15.0 cm is placed horizontally along the principal axis of a
converging lens of focal length 10.0 cm. If the closest end of the rod is 20.0
cm from the lens calculate the length of the image formed.

Solution

a) 1  ( nmaterial 1)( 1  1 ) b) 1 11
f nmedium R1 R2 f uv
1 11
1  (1.5 1)( 1  1 ) f uv 1 11
21.5 1 R1  1  1 1 10 35 v2
10 20 v1 v2  14cm
R1  10.8cm v1  20cm

Length of the image formed,

v1  v2  20 14  6cm

LO 7.3 Thin lenses Non-face-to face SLT

Exercise 144

An object is placed 16.0 cm to the left of a lens. The lens forms an image which is 36.0 cm to the right
of the lens.
a. Calculate the focal length of the lens and state the type of the lens.
b. If the object is 8.00 mm tall, calculate the height of the image.

Solution b) M   v  hi
u h
a) 1  1  1
f uv M   36  hi
1 1 1 16 0.8
f 16 36
f  11.08cm hi  1.8cm

Convex len

CHAPTER 8

PHYSICAL OPTICS

8.1 Huygen’s Principle
8.2 Constructive interference and destructive

interference
8.3 Interference of transmitted light through

double slits
8.4 Interference of reflected light in thin film
8.5 Diffraction by a Single slit
8.6 Diffraction Grating

MODE Face to face Non Face to face
SLT SLT
Lecture
Tutorial 2 2

8 8
(2 : pre &post lab)