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Chapter 26 Geometrical optics

26.1 Curved mirrors

(i) Concave mirrors

clip_image002

C:centre of curvature

F : focus

f : focal length

r :radius of curvature

CF = FX

FX = FP (P and X are close)

CF = FP

\clip_image004

(ii) Convex mirrors

clip_image006

C : centre of curvature

F : focus

f : focal length

r:radius of curvature

CP = CY

FY = FP (P and Y are close)

CF= FP

\ clip_image004[1]

Example 1

Rays parallel to the principal axis are incident on a concave mirror.

(a) Sketch a diagram showing the paths of the rays after reflection.

(b) Hence prove that the focal length of the mirror is half of its radius of curvature. State any assumption you have made.

26.2 Image Forming by Curved Mirrors

clip_image009

• Real

• Inverted

• Diminished

clip_image011

• Real

• Inverted

• Same size

clip_image013

• Real

• Inverted

• Enlarged

clip_image015

• Virtual

• Upright

• Enlarged

clip_image017

• Virtual

• Upright

• Diminished

26.3 Curve Mirror Equation

clip_image019

O: object

I : image

Since

clip_image021

clip_image023……….(1)

Also clip_image025

Then clip_image027……………..(2)

Since (1) =(2),

clip_image029

clip_image031

As X approaches P,

clip_image033

clip_image035 ( Formula for curved mirrors)

Convention:

Real is positive, virtual is negative

 

Sign of f

Concave mirrors

Positive

Convex mirrors

Negative

Example 2

A concave mirror has a radius of curvature of 40.0 cm.

(a) Sunlight parallel to the axis of the mirror is incident on the mirror. Where is the sunlight brought to a focus?

(b) An object of height 10 cm is placed 40 cm from the mirror.

(i) Find the position of the image,

(ii) Find the height of the image.

(iii) State the characteristics of the image.

(c) Sketch a diagram to show how the rays from the object form an image.

26.4 Refraction at curved surfaces

clip_image037

n1 sin i1 = n2 sin i2

Since i1 and i2 are small, if written in radian sin i1 » i1

n1 i1 = n2 i2

where

clip_image039

clip_image041

clip_image043

clip_image045

Rearranged the equqtion into

clip_image047

clip_image049 (As P is close to N)

clip_image051

Example 3

An object O is located 6 cm from a glass sphere with a diameterof 30 cm. The refractive index of the glass sphere is 1.50.Determine the nature and position of the image.

26.5 Thin lenses

clip_image053

For curved surface PP’

clip_image055

For curved surface QQ’

clip_image057

where v’ =virtual object distance

clip_image059

clip_image061

Lens maker formula:

clip_image063

clip_image065

If n1=1, n2=n

clip_image067

Example 4

A camera with a lens of focal length 10.0 cm is used to take a sharp picture of a distant landscape.

(a) What is distance of the lens to the film?

(b) The camera is then used to photograph a flower 50.0 cm away. How far, and in which direction, must the lens be moved to obtain a clear image on the film?

5 Telescopes and Microscopes

clip_image068

clip_image069

where

b= angle subtended at the eye by the image

a = angle subtended at the eye by the object

clip_image071 normal adjustment

For astronomical telescope;

Or

clip_image072

Chapter 25 Electromagnetic waves

25.1Electromagnetic vibrations

clip_image002

Electric vibration E = E0 sin (wt – kx)

Magnetic vibration B = B0 sin (wt- kx)

E, B and the direction of propagation of the electromagnetic wave is always perpendicular to one another

The speed of electromagnetic waves

clip_image004

Electromagnetic waves such as light are transverse waves. However, it is not particles which oscillate, but an electric field (E) coupled with a magnetic field B , as shown above. The speed of the waves through space, called the speed of light,c depends on the permittivity,e0 and permeability m0, of a vacuum :

speed of light (in vacuum) clip_image006 = 3.0 x 108 m s-1

25.2 Relationship between e0, m0 and c

The relationship of e0 , m0 and c

Velocity of electromagnetic wave, c = clip_image008

where

e0 = 8.85 x 10-12 F m-1 (Free space permittivity) and

m0 = 4p x 10-7 H m-1 (Free space permeability)

c = 2.998 x108ms-1

Example 1

The velocity of an electromagnetic waves in vacuum is given by the equation clip_image010

(a) Calculate this velocity.

(b) What deduction can be made from this information?

(c) What is it that oscillates in the electromagnetic waves?

25.3 Electromagnetic wave spectrum

The electromagnetic spectrum

Light Is one member of a whole family of transverse waves called the electromagnetic spectrum. In empty space, these waves all travel at the same speed: 300 000 km s-1 . Electromagnetic waves are emitted whenever electrons or other charged particles oscillate or lose energy. The greater the energy change, the lower the wavelength.

clip_image012

A table comparing electromagnetic wave and mechanical wave (e.g. sound wave)

Characteristic

E.M Wave

Mechanical Wave

Propagates through vacuum

Yes

No

Requires a medium for propagation

No

Yes

Speed

3.0 x 108 m s-1

(speed of light)

330 m s-1 (for sound)

Type of wave

Transverse

Longitudinal

(Sound waves)

and transverse

(Water waves)

4 Electromagnetic Wave Spectrum

Type

Wavelength/m

Frequency/Hz

Radio wave

> 1

< 108

Microwave

10-4 – 1

108 – 1012

Infrared

10-6 – 10-4

1012 – 1014

Visible light

4 x 10-7 – 7 x 10-7

75 x 1014 – 4.3 x 1014

Ultraviolet

10-9 – 10-7

1015 – 1017

X-rays

10-11 – 10-9

1017 – 1019

Y- rays

< 10-10

> 1018

Example 2

The wavelength of visible light ranges from 4.0 x 10-7 m to 7.5 x 10-7 m. What is the frequency range of visible light?

Chapter 24 Electronics

Electronic circuits handle small, changing electric currents. The changes are called signals. The diagram below shows the main parts of a basic electronic system.

clip_image002

Note:

•Devices which change signals from one form to another (e.g. sound to electrical) are called transducers(a device that converts one type of energy to another).

•Input signals can be very weak. But by means of electronic circuits, they can control output signals which can be much stronger. The output power is provided by the supply (usually low voltage DC).

clip_image004

Two possible functions of electronic circuits are described below.

1. Switching Depending on the input signals received, the output voltage is either HIGH (close to the supply voltage) – or LOW (zero), so the output device is either ON or OFF.

2. Amplification The output signals are an amplified (magnified) version of the input signals. For example, very low voltage AC from a microphone causes a higher voltage AC output for a loudspeaker.

The voltagegain of an amplifier is defined like this:

clip_image006

Note:

•In electronics, the term ‘voltage’ is commonly used for potential, PD, and EMF.

The key component in electronics is the transistor. It can amplify or act as a switch. An integrated circuit (IC), in a package as on the right, may have thousands of transistors and other components formed on a single chip of silicon.

Input devices

Input transducers (e.g. microphones) are called sensors. They must be linked to the circuit in such a way that any change they detect (e.g. a pulse of sound) causes a change in input voltage.

Sensors generating a voltage These include some microphones, and those light sensors that work like solar cells . Ideally, the voltage is in proportion to the external change causing it.

Sensors with varying resistance These include:

• light-dependent resistors (LDRs) ,

•thermistors (temperature-dependent),

•strain gauges (resistance changed by stretching).

To produce the necessary voltage change, the sensor forms one part of a potential divider . For example, when light falls on the LDR below, its resistance falls. The LDR therefore takes a smaller share of the supply voltage, so the voltage V falls.

clip_image008

Output devices

clip_image010

These include loudspeakers, buzzers, LEDs, and relays. Light-emitting diodes (LEDs) These emit light when a small current passes through. They can be used as indicator lamps to show the presence of an output voltage. Like all diodes, they only conduct in one direction (see D1 and D2).

To avoid damage, a resistor must be placed in series with an LED to limit the current through it.

The LED on the right can take a current of 0.01 A, which produces a 2 V drop across it. If a circuit’s output voltage is 9 V, there needs to be a 7 V voltage drop across the resistor, so its resistance should be 7/0.01 = 700 S2.

Relays These are electromagnetic switches. A small current activates an electromagnet. This closes (or opens) a switch which can control the flow of a larger current in a separate circuit, e.g. a circuit with a mains heater or motor in it.

clip_image012

Impedances and matching

Resistance is one form of impedance (see D4). The microphone and loudspeaker below each have impedance. The amp[ifer has input impedance and output impedance rather as a battery has internal resistance .

clip_image014

In the diagram, the various impedances are represented by resistors in the input and output circuits.

In each circuit, it is the higher impedance that has the higher voltage drop across it. So

For maximum input voltage, the microphone’s impedance should be as low as possible.

For maximum output voltage, the loudspeaker’s impedance should be as high as possible. However, this gives a very low output current and almost no sound!

For maximum output power, the loudspeaker’s impedance should be the same as the output impedance. (Output voltage x output current is then at its highest.)

A transformer can be used to match impedances and give maximum power transfer. A transformer with an impedance Z in its primary circuit is equivalent to an impedance of Z(N2/ N1)2.

24.1 Operational amplifiers (Open Loop)

Op Amp beroperasi asas untuk menggandakan voltan.

Open loop bermaksud bila di bekalkan Vin di dalam op amp ianya akan menghasilkan Vo, voltan output.

Di bawah di tunjukkan gambarajah op amp

clip_image016

This IC has one output and two inputs. It amplifies the different( between the two input voltages. It is a differential amplifer.

• If V+ > V the output voltage is positive (+).

• If V+ < V the output voltage is negative (-).

• If V+ = V the output voltage is zero.

If only one input is in use (and the other is at zero voltage),

• a positive (+) voltage on the inverting input causes a negative (-) output voltage,

• a positive (+) voltage on the non-inverting input causes a positive (+) output voltage.

Note:

• The + and – signs on the inputs show whether they invert of not. They do not indicate + or – voltage.

Power supply A three-terminal DC supply is required, giving 0 V (earth) and typically, ±9 V. The supply connections are often excluded from diagrams.

Op amp features These include:

• an extremely high voltage gain, typically around 105, though less at high frequencies,

• a very high input impedance (e.g. 1012 W),

• a very low output impedance (e.g. 102 W).

The above op amp is also known as a differential Amplifier- it amplifies the voltage between Vin (non-inverting and inverting voltage), so the Vo given by

clip_image018

A0, ialah open loop gain (pemalar gandaan)

clip_image020

Graf Hubungan diantara Vo vs Vin (V+ – V)

clip_image022

Daripada graf didapati

1. Gradient ,m = A0

2. Op amp tetap juga beroperasi walaupun A0 mengganda sehingga 105 kali, tetapi inya bergantung kepada voltan yang di bekalkan.

3. voltan yang di gandakan tidak boleh melebihi voltan yang dibekalkan, pada kaadaan ini ianya dinamakan voltan tepu.

Contoh 1

clip_image024

The circuit diagram shows an op-amp with an open loop gain of 105 connected to a ±9V voltage supply. Sketch graph to show the variation of the input voltage Vin and output voltage Vout with time t for each of the following input voltages.

(a) sinusoidal a.c voltage of peak value 20mV

(b) sinusoidal a.c voltage of peak value 120 mV

(c) sinusoidal a.c voltage of peak value 2.0 V

24.2 Op amp (close loop)

Differential amplifier – it amplifies the voltage between inverting and non inverting, so V output given by

clip_image018[1]

Open loop gain (gandaan voltan gelung-buka), clip_image027

In the case of op amp, A0 is depend on the frequency.

To reduce gain(actually to reduce frequency), op amp are used with a resistor or wire linking the output to one input.

clip_image029

This kind of circuit is called a close loop.

Close loop bermaksud sebahagian V0 disambung secara selari kepada voltan Vin (contohnya kepada V.

When connecting output to the inverting input. The connection is called with another name as

Negatif feed close loop back in short it just called a Negatif feedback.

Negative feedback

clip_image031

The amplifier circuit on the right uses a closed loop to feed back a set fraction of the output voltage to the inverting input. This is called negative feedback because the signal being fed back partly cancels the input signal.

Advantages using negatif feeback

  1. voltan output tidak sefasa dengan voltan input
  2. mengurangkan voltan input setelah ianya disuap balik
  3. reducing gain (voltage gain)

Advantages of reducing gain (Using closed loop op amp) These include the following:

• Input signals can be stronger without causing saturation and, therefore, distortion.

• The gain is less affected by temperature changes.

• The bandwidth is greater, i.e. the gain is constant over a wider range of input frequences (see below).

Note : Bandwidth = frequency which the gain is constant

Graf A0 vs frekuensi

clip_image033

24.3 Inverting and non-inverting amplifiers

V0 = A0 (V2 – V1) where A0= open loop gain

1 Inverting amplifier

Gain clip_image035

clip_image036

Explanation:

Op amp as an inverting amplifier

clip_image038

The amplifier circuit on the right uses a closed loop to feed back a set fraction of the output voltage to the inverting input. This is called negative feedback because the signal being fed back partly cancels the input signal.

To calculate the amplifier’s closed-loop voltage gain clip_image040 the following assumptions are made:

• The op amp is not saturated. It has such a high open-loop gain that the voltage difference between its inverting and non-inverting inputs must be negligible. Therefore P is effectively at 0 V. (P is called a virtual earth.)

• The current through P is negligible. Therefore, the current through Rf is the same as through Ri (see D3).

If I is the current through Rf (feedback resistor) and Ri,

voltage drop across Rf = IRf

voltage drop across Ri = IRi

But if P is at 0V,

voltage drop across Rf = 0 – Vout = -Vout

voltage drop across Ri = Vin – 0 = Vin

From the above, it follows that clip_image042

So closed-loop voltage gain = clip_image044

From the closed-loop voltage gain equation below:

• An inverting amplifier has a negative gain.

• The gain does not depend on the characteristics of the op amp. It is set by the values of Rf and Ri. For example, if Rf = 200 kW and Ri = 20 kW, the gain = -200/20 = -10.

Contoh 2

The circuit diagram shows an inverting amplifier the power supply is ± 9V.

clip_image046

(a) what is the voltage gain of the circuit ?

(b) what is the output voltage V0.

(i) x = 200 kW and Y = 50 kW

(ii) x = 50 kW and Y = 150 kW

2 Non-inverting amplifier

Gain clip_image048

clip_image050

Explanation:

Op amp as a non-inverting amplifier

clip_image052

In the circuit on the top, the input signal goes to the non-­inverting input, and the potential divider provides negative feedback. With the same assumptions as above,

closed-loop voltage gain clip_image054

A non-inverting amplifier has a much higher input impedance than the inverting type. This is useful, for example, where a high-impedance microphone is connected to the input (see Impedances and matching in H16).

Voltage follower This is a non-inverting amplifier in which Rf is zero (i.e. a direct connection) and Ri is infinite (i.e. removed). In this case, the gain is 1, so the output voltage is the same as the input voltage. However, the output impedance is much less than the input impedance, which makes the circuit useful for impedance matching.

Contoh 3

In the circuit diagram shown, the input signal is Vin and the output signal is V0

clip_image056

(a) state and explain the type of amplifier

(b) what is the output voltage V0 if Vin

(i) +1.5 V

(ii) +2.5 V

(c) sketch a graph to show the variation of the output voltage V0 with time t, if

clip_image058

24.4 Use of operational amplifiers

3 Uses of op-amps

(a) Changing sinusoidal wave forms to rectangular waveforms

clip_image060

(b) Automatic electrical switch/ Comparator

clip_image062

On the right, the LED is switched on automatically when darkness falls. The switching is done by an op amp which compares the voltages from two potential dividers:

V is fixed by the ratio of R2 to R1,

V+ varies with the resistance of the LDR.

There is no feedback, so a small difference between V and V+ saturates the op amp, causing maximum output voltage.

In bright light, the LDR has a low resistance, so V+ is low and less than V . The op amp is saturated, but its output voltage is negative, so the LED cannot conduct.

As the light intensity falls, the resistance of the LDR rises, and so does V+. When V+ > V the op amp saturates in the opposite direction. The output voltage is now positive, so the LED conducts and lights up.

(c) Voltage follower

clip_image064

V0=Vi, Gain ACL = 1

(d) Summing amplifier

clip_image066

clip_image068

(e) Differential Op-Amp

clip_image070

clip_image072

(f) AC Amplifier

clip_image074

(g) The Schmitt Trigger

clip_image076

24.5 Oscillators

clip_image078

The Wein oscillator – used both positive and negative feedback

1.positive feedback through R1 and C1

2.Negative feedback through resistor in series

3. R1=R2 = R, C1 = C2 = C

4. Frequency clip_image080

Chapter 23 Alternating currents

AC terms

The graph below shows how the current from an AC supply varies with time. Here are some of the terms used to describe AC. Note the similarities with those used for circular motion and SHM.

Continue reading

Chapter 22 Electromagnetic induction

 

clip_image002

Above, a magnet is moved into a coil. If the flux through the coil changes (at a steady rate) by DF in time Dt, then an EMF of clip_image004 is induced in each turn. But there are N turns in series. So, the total induced EMF E is as follows:

clip_image006

For example, if the flux changes by 6 Wb in 2 s, and the coil has 100 turns, then the total induced EMF is 300 V.

Continue reading

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