Physics 72.1 Practical Exam Reviewer

PHYSICS 72.1 REVIEW FOR PRACTICAL EXAM 2ND SEMESTER, A.Y. 2015-2016 R. AGUILAR, N. CABELLO, D. LUMANTAS NATIONAL INSTITUTE OF PHYSICS UNIVERSITY OF THE PHILIPPINES DILIMAN, QUEZON CITY 1101

ELECTRIC FIELD & EQUIPOTENTIAL LINES

ELECTRIC FIELD & EQUIPOTENTIAL LINES

Illustration of the difference between electric field vectors and electric field lines. Reference: Physics 72.1 Electric Field & Equipotential Lines Lab Manual 2015

ELECTRIC POTENTIAL DIFFERENCE  Δ𝑉: potential difference  𝑊𝐴→𝐵 : work done on a positive test charge qo as it moves from A to B

ELECTRIC FIELD & EQUIPOTENTIAL LINES Equipotential Lines (green)

 Points in space that have the same

electric potential with respect to the same reference point

Electric Field Lines (red)

 Always perpendicular to an

equipotential line (see Eq. 4)

 Electric field points towards

decreasing potential

[1] http://www.alpcentauri.info/equipotential_lines.html

Equipotential and electric field lines of 2 equal but oppositely charged particles [1]

1) WHICH POINT HAS THE LARGEST MAGNITUDE OF ELETRIC FIELD? 2) WHAT IS THE DIRECTION OF ELECTRIC FIELD AT THAT POINT?

D

A

B

C

ANSWERS: 1) A. Equipotential lines are closest at that point. 2)  (to the right). Since electric field lines are always perpendicular to equipotential lines and they point towards decreasing potential.

OHM’S LAW

OHM’S LAW V:Voltage (V) R: Resistance (Ω) I: Current (A) ρ: resistivity (Ωm)

OHM’S LAW PLOTS (y vs. x) 1) Voltage vs. Current  Slope = R

2) Current (y) vs. 1/Resistance  Slope = V

3) Voltage vs. Resistance  Slope = I

4) Resistance vs. Length  Slope = ρ/A

OHM’S LAW (LINEAR REGRESSION) Current (A) Voltage (V)

Example 1 (Voltage vs. Current)

0.1

0.3

0.2

0.7

1.8

0.3

1.09

1.6

0.4

1.4

0.5

1.82

Voltage vs. Current

2

Voltage (V)

1.4 y = 3.9x - 0.084 R² = 0.9994

1.2 1

3) USING THE GIVEN VALUES AND THE PLOT, WHAT IS THE VALUE OF RESISTANCE OF THE MATERIAL?

0.8 0.6 0.4 0.2 0 0

0.1

0.2

0.3 Current (A)

0.4

0.5

0.6

Answer: Based from the slope of the graph, 𝑹 = 𝟑. 𝟗 𝛀.

OHM’S LAW (LINEAR REGRESSION) Length (cm) Example 2 (Resistance vs. Length of Wire)

100 120 140 160 180 200

Resistance vs. Length of Wire 3.5

Resistance (Ω)

3 2.5 2

1 0.5 0 0

50

100 150 Length (cm)

200

2.1 2.3 2.48 2.65 2.83 3.07

4) USING THE GIVEN VALUES AND THE PLOT, IF THE CROSS-SECTIONAL AREA IS 2 cm2, WHAT IS THE RESISTIVITY OF THE MATERIAL?

y = 0.0094x + 1.1552 R² = 0.9973

1.5

Resistance (Ω)

250

Answer: 𝜌 = 𝑠𝑙𝑜𝑝𝑒 ∗ 𝐴 = 0.0094 * 2 𝝆 = 𝟎. 𝟎𝟏𝟖𝟖 𝛀𝐜𝐦

CIRCUIT ANALYSIS

CIRCUIT ANALYSIS  Circuit – conducting path where current can flow and the components that make up

this path.  Steady current – only possible for closed loops or complete circuits with at least one source of

electromotive force (emf) that supplies electrical energy to the circuit.  For circuits composed of resistors connected in both series and/or parallel, Ohm’s law applies:

𝑽 = 𝑰𝑹𝒆𝒇𝒇 where Reff is the effective resistance of the circuit

CIRCUITS IN SERIES AND PARALLEL

 In Series:

𝑅𝑒𝑓𝑓 = 𝑅1 + 𝑅2 + 𝑅3 + …+𝑅𝑁

 In Parallel:

1

𝑅𝑒𝑓𝑓

1 1 1 1 = + + +⋯+ 𝑅1 𝑅2 𝑅3 𝑅𝑁

EXAMPLE CIRCUIT – FROM CLASS

R2 and R3 are in parallel (R2||R3) R1 is in series with R2||R3

1 1 𝑅𝑒𝑓𝑓 = 𝑅1 + 𝑅2 ||𝑅3 = 𝑅1 + + 𝑅2 𝑅3

−1

𝑅2 𝑅3 = 𝑅1 + 𝑅2 +𝑅3

KIRCHOFF’S RULES  Loop Rule

 Junction Rule

 σ𝑖 𝑉𝑖 = 0

 σ 𝐼𝑒𝑛𝑡𝑒𝑟𝑖𝑛𝑔 = σ 𝐼𝑙𝑒𝑎𝑣𝑖𝑛𝑔

 Sum of changes along a closed path is zero

 Current in = current out

𝑉1 − 𝐼𝑅1 − 𝐼𝑅2 − 𝑉2 − 𝐼𝑅3 = 0

𝐼1 = 𝐼2 + 𝐼3 + 𝐼4

http://www.wikipremed.com/01physicscards.php?card=708

5) SET-UP JUNCTION RULE AT PT. P AND LOOP RULE FOR A AND B.

I1

P I3

A

I2 B

Answers:

Loop A: 𝑉1 − 𝐼1 𝑅1 − 𝐼2 𝑅2 = 0 Loop B: 𝑉2 − 𝐼2 𝑅2 = 0 Junction P: 𝐼1 + 𝐼3 = 𝐼2 : Outer Loop (redundant with Loops A and B): 𝑉1 − 𝐼1 𝑅1 − 𝑉2 = 0

MEASURING PARAMETERS USING MULTIMETER Measuring voltage across resistor Measuring current through circuit

Measuring resistance of resistor

CAPACITORS AND RC CIRCUITS

CAPACITOR  Device that stores electrical energy

Capacitance: 𝑞 𝐶= 𝑉 𝑞 −Charge stored in the capacitor 𝑉 − Potential difference between the capacitor

https://en.wikipedia.org/wiki/Capacitor

RC CIRCUIT Discharging:

Charging:

Time constant: 𝜏 = 𝑅𝐶

6) WHAT IS THE TIME CONSTANT OF THE RC CIRCUIT BELOW? 7) WHAT IS ITS VOLTAGE AT t = 4RC? CHARGING

100 𝑀Ω

5.5V

1000 𝑛𝐹

Answers: Time constant 𝜏 = 100𝑥106 Ω ∗ 1000𝑥10−9 F 𝝉 = 𝟏𝟎𝟎 𝒔 Voltage at t=4RC 𝑉 𝑡 = 𝑉𝑜 1 − 𝑒 −𝑡/𝑅𝐶 𝑉 = (5.5 𝑉) 1 − 𝑒 −4𝑅𝐶/𝑅𝐶 𝑉 = 5.5𝑉 1 − 𝑒 −1 𝑉 = 0.63 5.5 𝑉 𝑽 = 𝟓. 𝟑𝟗𝟗 𝑽 = 𝟓. 𝟒𝟎 𝑽

ELECTROMAGNETIC INDUCTION

MAGNETIC FLUX Magnetic flux changes by changing:  the magnitude of the

magnetic field  changing the surface area  changing the relative

  B  B  A http://ibphysicsstuff.wikidot.com/electromagnetic-induction

orientation of the field and the surface normal Change in flux

Induced emf

FARADAY’S LAW OF INDUCTION  The induced emf in a closed loop equals the negative of the

time rate of change of the magnetic flux through the loop

Lenz’s Law:

d B   dt

An induced current will be in such a direction as to produce a magnetic field that will oppose the motion of the magnetic field that is producing it.

INDUCTION EXPERIMENT 1 Magnet

Iinduced

Binduced

North pole moves toward solenoid

cw

right

South pole moves towards soleniod

ccw

left

ccw

left

cw

right

North pole moves away from solenoid South pole moves away from solenoid * as viewed from left to right

http://voer.edu.vn/c/faradays-law-of-induction-lenzs-law/0e60bfc6/99a3eaad

Actual Direction

INDUCTION EXPERIMENT II  Magnetic Field of a solenoid

𝐵 = 𝜇𝑛𝐼  Magnetic permeability of air and aluminum: almost equal to 𝜇0  Magnetic permeability of iron > 𝜇0 Magnetic permeability

Magnetic Field

Greater deflection upon turning on and off

INDUCTION EXPERIMENT III

overlap

Increasing deflection upon turning on and turning off

8) WHAT IS THE MAGNETIC PERMEABILITY FOR N = 50 TURNS OF A 1.0-METER SOLENOID WITH THE FOLLOWING PLOT OF B VS. I? Current (A)

Magnetic Field (B)

0.1

0.3

1.8

0.2

0.7

1.6

0.3

1.09

0.4

1.4

0.5

1.82

Magnetic Field vs. Current

Magnetic Field (T)

2

1.4 y = 3.9x - 0.084 R² = 0.9994

1.2 1 0.8

Answer:

0.6

𝑁 50 𝑡𝑢𝑟𝑛𝑠 𝑛= = 𝐿 1.0 𝑚

0.4 0.2 0 0

0.1

0.2

0.3 Current (A)

0.4

0.5

0.6

𝝁 = 𝒔𝒍𝒐𝒑𝒆/𝒏 = 𝟎. 𝟎𝟕𝟖 𝑻𝒎/𝑨.

SOURCES OF MAGNETIC FIELD

SOURCES OF MAGNETIC FIELD  A wire carrying current produces a magnetic field (direction determined by right-hand rule)  When the wire is looped, the field near the center becomes perpendicular to the direction of the loop  Multiple loops increase the field strength – solenoid

𝜇0 𝐼 𝐵= 2𝜋𝑟

Wire is looped

Multiple loops

𝜇0 = 4𝜋 × 10−7 𝑇 ∙ 𝑚/𝐴 Magnetic permeability of vacuum

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/solenoid.html#c1

𝜇0 𝑁𝐼 𝐿 𝐵 = 𝜇0 𝑛𝐼

𝐵=

SOURCES OF MAGNETIC FIELD  Addition of material inside solenoid modifies magnetic permeability (µ0 becomes

µ=kµ0) 

Increase in magnetic permeability results to increase in B strength

http://physics.stackexchange.com/questions/90609/1-tesla-electromagnet

MAGNETIC FIELD LINES FROM DIFFERENT CONFIGURATIONS Single bar magnet

Two bar magnets unlike poles facing each other

Solenoid

Two bar magnets like poles facing each other

INTERFERENCE AND DIFFRACTION

SUPERPOSITION OF WAVES  Consider two waves travelling through the same medium at the same time.

The net displacement of the medium at any point in space or time, is simply the sum of the individual wave displacements Interference: combination of two or more waves to form a composite wave

INTERFERENCE  Constructive and Destructive Interference

DIFFRACTION  Bending of waves as they pass by some objects or through an aperture

SINGLE SLIT EXPERIMENT 𝑎 sin 𝜃 = 𝑚𝜆 sin 𝜃 =

𝑦𝑚,𝑑𝑖𝑓𝑓 𝐿

𝑦𝑚,𝑑𝑖𝑓𝑓 a = 𝑚𝜆 𝐿

𝑎 – slit width 𝑦𝑚,𝑑𝑖𝑓𝑓 − mth intensity minimum 𝐿 − slit to screen distance 𝜆 − wavelength of the light source

(1) (2) (3)

DOUBLE SLIT INTERFERENCE

y

Condition for maximum: 𝑚𝜆𝐿 𝑦𝑚,𝑖𝑛𝑡 = 𝑑

Diffraction envelope

𝑎 – slit separation 𝑦𝑚,𝑖𝑛𝑡 − mth intensity peak from the center 𝐿 − slit to screen distance 𝜆 − wavelength of the light source

9) GIVEN THE FOLLOWING FIGURE, WHAT IS THE WAVELENGTH OF THE LIGHT SOURCE? Better to use single-slit diffraction equation since the given Δ𝑦𝑚 is at the dark fringes (corresponding to 𝑚 = 2) of the diffraction pattern. Thus, 𝑦𝑚,𝑑𝑖𝑓𝑓 𝑎 =𝜆 𝑚𝐿

a LIGHT SOURCE

d

Δy𝑚 = 4 mm

a

0.04 𝑚𝑚

4 𝑚𝑚 ( 2 ) 2 ∗ 1.0 𝑚

=𝜆

𝜆 = 40 𝑛𝑚

L = 1.0 m

a = 0.04 mm d = 0.25 mm

Note: This wavelength is not within the visible range of light (just placed random values).

OPTICAL DISK – REFRACTION & REFLECTION

THE SPEED OF LIGHT

 Light slows down when travelling in a medium other than air/vacuum  The ratio between the speed of light in vacuum (c) and its speed in some medium (v) is given by

𝑐 𝑛= , 𝑣

𝑐 =3×

108

𝑚 𝑠

n is called the index of refraction  Light travels faster in vacuum/air compared to other media; v is always less than c; n has value > 1

LAWS OF REFLECTION AND REFRACTION Note: Law of reflection holds for all types of mirrors, i.e plane and spherical mirrors.

 Law of reflection:

𝜃1 = 𝜃1′  Law of refraction (Snell’s Law)

1 1 sin 𝜃1 = sin 𝜃2 𝑣1 𝑣2 𝑛1 sin 𝜃1 = 𝑛2 sin 𝜃2  Total internal reflection

𝜃𝑐 =

sin−1

𝑛2 𝑛1

(Special case of Snell’s Law where 𝜃2 = 90°; no light is refracted at angles greater than 𝜃𝑐 )

REFLECTION IN SPHERICAL MIRRORS

Concave

Convex

RAY TRACING FOR DIFFERENT REFRACTING MEDIA

CYLINDRICAL LENS 𝜃𝑖 : angle of incidence 𝜃𝑟 : angle of reflection 𝜃𝑟 ’: angle of refraction 𝜃𝑖

𝜃𝑖

𝜃𝑟 Air

Glass

Air

𝜃𝑟

𝜃𝑟 ’

Glass 𝜃𝑟 ’

Refracted ray bends away from the normal since nglass > nair

Refracted ray bends towards the normal since nair < nglass

10) AT WHICH OF THE GIVEN SET-UPS DOES REFRACTION OCCUR AT THE AIR-TO-GLASS INTERFACE?

10) AT WHICH OF THE GIVEN SET-UPS DOES REFRACTION OCCUR AT THE AIR-TO-GLASS INTERFACE?

Answer: Hint: Draw a normal line at the air-to-glass interface. Refraction only occurs at C.

POLARIZATION – MALUS’ LAW

POLARIZATION – MALUS’ LAW

POLARIZATION – MALUS’ LAW Eo: amplitude of the incident electric field θ: angle between the polarization of the incident light and the transmission axis Itrans: intensity of transmitted light

𝐸𝑡𝑟𝑎𝑛𝑠 = 𝐸𝑜 cos(𝜃) 2 𝐼𝑡𝑟𝑎𝑛𝑠 = 𝑘𝐸𝑡𝑟𝑎𝑛𝑠 = 𝑘𝐸𝑜2 cos 2 𝜃 = 𝐼𝑜 cos 2 𝜃

11) GIVEN THE FFG. CONFIGURATION, WHERE SHOULD A THIRD POLARIZER BE PLACED FOR THE LIGHT SENSOR TO HAVE AN INTENSITY READING GREATER THAN 0 LUX? 12) WHAT ARE THE ALLOWED ANGLES?

A

B

C 0 lux

sensor

𝜃 = 00

𝜃 = 900

Answers: B. In between the two polarizers. Allowed angles: 00 < 𝜃 < 900

CONCEPTS

 Laser diode – linearly polarized light  Plain light source – not linearly polarized

Intensity of transmitted laser diode light source changes as polarizing angle is varied  Malus’ Law  Obeyed for both laser diode and plain light source (as seen on data for the two polarizers)

CONCEPTS  Intensity of light source does not affect Malus’ Law behavior.  Both laser diode and plain light source plots exhibit Malus’ Law as seen on

𝐼𝑒𝑥𝑝/𝐼𝑚𝑎𝑥 vs. 𝜃 plots

 Transverse nature of EM waves is shown with the reduction of intensity

as angle of polarization changes.  No more light is transmitted when the angle of polarization is completely

perpendicular to the direction of propagation.

IMAGE FORMATION

THIN LENS EQUATION Lens equation:

1 𝑠𝑜

+

1 𝑠𝑖

Linear magnification: 𝑀

Parameters: 𝑠𝑜 − 𝑜𝑏𝑗𝑒𝑐𝑡 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑠𝑖 − 𝑖𝑚𝑎𝑔𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓 − 𝑓𝑜𝑐𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ ℎ𝑜 − 𝑜𝑏𝑗𝑒𝑐𝑡 ℎ𝑒𝑖𝑔ℎ𝑡 ℎ𝑜 − 𝑖𝑚𝑎𝑔𝑒 ℎ𝑒𝑖𝑔ℎ𝑡

http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/lensdet.html

=

1 f

=−

𝑠𝑖 𝑠𝑜

=

ℎ𝑖 ℎ𝑜

Criteria for formed image: |M| > 1 : magnified |M| < 1 : reduced |M| = 1 : same height +M : upright -M : inverted

Positive object distance

real object in front of the lens (incident side)

Negative virtual object at the back of the lens (transmission side)

image distance

real image at the back of the lens (transmission side)

virtual image in front of the lens (incident side)

focal length

converging/convex lens

diverging/concave lens

COMBINATION OF TWO LENSES (IN CONTACT) Effective focal length (in contact): Concave lens (diverging): negative focal length (-f) Convex lens (converging): positive focal length (+f)

COMBINATION OF TWO LENSES (NOT IN CONTACT)



13) GIVEN THE FOLLOWING PARAMETERS, WHAT IS THE MAGNIFICATION OF THE IMAGE? Answer:

o = 5 cm f = 4 cm

1 1 1 + = 𝑜 𝑖 𝑓 1 1 1 + = 5 𝑐𝑚 𝑖 4 𝑐𝑚 𝑖 = 20 𝑐𝑚 𝑖 20 𝑀 = − = 𝑜 5 𝑀 =4

END