Module 4 + 6 Model Answers Flashcards
(92 cards)
Using Oscilloscope to measure Frequency
Oscilloscope to measure frequency
- Use sig. generator/tuning fork with mic and CRO
- Use loudest amplitude sound possible
- Adjust time base to get largest trace possible
- Adjust volts/div to get largest amplitude
- Multiply ms/div by number of divisions to get period - Frequency = 1/T
Demonstrating/Explaining interference patterns
- Use appropriate detector to measure intensity/amplitude
- Maxima = highest intensity, constructive interference, in phase, path difference = nλ, Phase difference = even integer of π
- Minima = lowest intensity, destructive interference, in antiphase, path difference = (n+1/2)λ, Phase difference = odd integer of π
I-V Characteristics PAG
- Set up circuit with variable resistor in series with component.
- Use variable resistor to change distribution of pd across circuit - record I at each setting.
- Plot graph of I against V
Formation of standing waves (eg string, sound, microwaves)
- Incident wave reflects off end of string/surface
- Reflected wave superposes/interferes with incident wave
- Antinodes - where always constructive and Amplitude = MAX
- Nodes - where destructive and displacement always ZERO
Finding Speed of sound from column of air
- Place tube vertically in bucket of water.
- Use signal generator to produce sound of known frequency and mic linked to CRO to detect
- Raise tube from 0 until Amplitude on CRO reaches max.
- This is first harmonic and so length is
- Multiple length by 4 to get
- Use c=3x10^8
Describing example of PE effect
- Gold leaf electroscope - Charge it up and gold deflects.
- Shine normal light in Zn disc & gold leaf remains deflected regardless of time.
- Shine UV & g. leaf immediately returns loss of charge from Zn
Explaining PE effect
- Electron emitted from surface of metal when light incident on surface
- Only occurs if frequencey > threshold frequency or if hf > work function of metal
- FTh in minimum f required to cause PE emission
- Work function is minimum energy required to remove photo electron from surface of metal.
- Rate of emission is proportional to Intensity (as long as if f > FTh)
Evidence as particular nature of light
- Were light a wave light of any f would eventually cause PE emission
- Changing intensity has no effect on emission
Conditions for interference with light
- Coherent sources (from a single monochromatic light split through slits),
- Narrow slits (to cause diffraction),
- Slits close together (so diffraction patterns can overlap),
- A distant screen (so the pattern is visible and measurable).
Double slit method to measure wavelength and limitations
Measure D with ruler, fringes over several for accuracy.
Use micrometer for tiny slit separation.
Main issue: % uncertainty in fringe width — measure n fringes.
Potential Dividers - Thermistor
- As T↓ R(thermistor)↑ so V(therm)↑ and V(fixed)↓
- To turn component ON as T↑ place across FIXED
- To turn component OFF as T↓ place across THERMISTOR
- Vice Versa for response to decreasing Temperature
Potential Dividers - LDR
- As L↑ R(LDR)↓ so V(LDR)↓ and V(fixed) ↑
- To turn component ON as L↑ place across FIXED
- To turn component OFF as L↑ place across LDR
- Vice Versa for response to decreasing Light
Describing and explaining interference patterns (eg walking past two speakers)
- Interference pattern is described by areas of Maxima and minima
- ·Maxima (loudest/brightest) - constructive interference - waves in phase - path difference 0 or any whole number of wavelengths nλ
- Minima (quietest/dullest) - destructive interference - waves in ANTIphase - Path difference = (n+1/2)λ
Young’s double slit as evidence of wave nature of light
- Light passed through single slit - then to double slit to create coherent waves
- Double slit produced interference - we know that water waves interfere - therefore light must be wave
Polarisation of Waves
- Can only occur with transverse waves
- Unpolarised wave has oscillations in all planes
- Polarising filter only allows through oscillations in direction of filter
Proving a light wave is polarised (eg reflected off water)
- View wave through filter
- Rotate filter and observe change in intensity
- When filter is at 0 degrees to plane of polarisation I = max
- When filter is at 90 degrees to plane of polarisation I = 0
- When filter is at 180 degrees to plane of polarisation I = max
Proving microwaves are polarised
- Use Microwave emitter and microwave receiver
- Either place grill/grating between and rotate OR
- Rotate receiver - Same as last 3 points above
Plank Constant PAG + limitations
Method:
- Use a bank of LEDs (known wavelengths) with a flying lead and variable resistor
- Gradually increase voltage until each LED just begins to glow
- Record striking voltage V
- Plot graph of V against f
- Gradient = h/e so h = gradient x e
Limitation:
- External light makes LED appear to glow early → striking V too low → value of h too high
- Solution: View LEDs through a dark tube to reduce ambient light
Describe and explain how the capacitor plates A and B acquire opposite charges when the switch is closed
Electrons in the circuit move in a clockwise direction and electrons are deposited on plate B. An equal number of electrons are removed from plate A giving it a positive charge of equal magnitude.
State and explain what affects the time constant of a capacitor-resistor circuit.
- Double the resistance and time constant doubles and the rate of flow of charge halves.
- Double Capacitance doubles time constant as twice a much charge is needed for each volt increase or decrease.
A student investigates the discharge of a capacitor.
Explain what graph they would plot to obtain a linear relationship and what key values can be obtained from the graph.
V=V₀eᵗ/ᶜᴿ
lnV=-t/CR
lnV=(-1/CR)t + lnV₀
Gradient (dy/dx)=-1/CR
so cr=-1/gradient.
The capacitor consists of two parallel metal plates separated by air. The switch is closed to charge the capacitor. The switch is then opened and the separation between the charged plates is doubled. State and explain what happens to the energy stored by the capacitor.
The charge on each plate remains the same. C = ε0A/d, hence the capacitance is halved. E = ½ Q2/C, E ∝ 1/C and hence energy stored doubles.
The capacitance of B is twice the capacitance of A. Explain why the potential difference across capacitor A is twice the potential difference across capacitor B.
The charge Q is the same on each capacitor is the same V = Q/C so V ∝ C⁻¹ hence CA is half CB so VA will be double VB
Describe how a student could investigate how the voltage varies with time when a capacitor is discharged. PAG
- Choose values of C and R to give a time constant of about 2 minutes (Allows time to measure)
- Close switch to charge up capacitor. Measure initial p.d. and open the switch and recorded V and time every 20s for 5 minutes. Plot a graph of V against time. (or InV against t gradient = - CR)