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Question from Teacher Joshua: changes during refraction?

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Question from Teacher Joshua: changes during refraction?

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When light passes across a boundary into water or glass in undergoes refraction. Refraction doesn’t change the colour of light, but its wavelength does change during refraction.

Explain how is this possible.

Hint: 1. What stays constant during refraction and what changes?

2. Think about topic E1 atomic physics and the cause of colour.

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Why are some isotopes more stable than others? 

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Why are some isotopes more stable than others? 

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The neutron to proton ratio increases at a rate higher than the linear (A-Z) / Z = 1. This is because as the nucleus gets larger, there needs to be a corresponding increase in the attractive strong nuclear force, and adding neutrons manages to fulfill this. Protons cannot be used as while the strong nuclear force increases, so does the repulsive electrostatic force, leading to an unstable nucleus.

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Radioactive waste decays in the order of alpha or beta minus / plus decay, then gamma decay. All types of radiation are dangerous, but gamma radiation are the most dangerous to living organisms as they have the greatest penetrative ability of the different types of radiation, while alpha and beta radiation aren't able to penetrate human skin and tissue as easily. While gamma radiation's ionizing power is not as strong as the other types of radiation, the energy it releases is immense and will still be very harmful to living organisms when exposed, damaging the DNA structure and causing cancer cells to form and develop into tumors.

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When fission happens, the neutron is first absorbed by the atom before the nucleus (with the atom + the neutron) splits apart.

Mass of product - mass of reactants = mass defect

Mass defect is difference in mass converted into energy

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Beta decay graph being continuous is proof for the existence of anti-neutrinos and neutrinos

idk dawg

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Nuclear density is approximately constant as nucleon number increases. This is because volume is proportional to nucleon number A via:

R = R₀A^(1/3)

So radius (R) is proportional to A (atomic number)

Volume of a sphere has r^3

Density = mass/volume

density is proportional to 1/r^3 which is proportional to 1/A

therefore mass is proportional to A

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DP2 Unit E Summative HL WORKED SOLUTIONS (Take Action)

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Ans: B

Remember that mass of the sum of all the particles in a nucleus is greater than the actual mass of the nucleus

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Continuous means that the line in the graph doesn't have a gap in it

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Current goes through induction coil -> coil is made out of conductive metal/material -> Magnetic field is created -> Magnetic field induced by coil goes from down to up -> Other coil gets electrons transferred from the wireless charger induction coil

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Derived expression for magnetic flux as a function of time

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Derived expression for magnetic flux as a function of time

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Torque = -kw = I x alpha = Frsin(theta)

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Q1. Right-hand‐rule description and operation of the rail-gun diagram

1. Current path: A DC supply drives current up one rail, across the conducting armature (projectile), and back down the other rail.

2. Magnetic field direction: By the right‐hand‐grip rule, each rail carries current that produces circular magnetic field lines around it; between the rails these fields add to produce a strong field directed into (or out of) the plane depending on current direction.

3. Lorentz force on armature: The armature carries current perpendicular to the magnetic field. Applying the right‐hand rule for the motor effect (thumb = force, forefinger = field, middle finger = current) shows a force along the rails pushing the projectile forward.

Q2. Definition of electromagnetic induction
Electromagnetic induction is the process by which a changing magnetic flux through a conductor induces an electromotive force (emf) and hence a current in that conductor (Faraday’s law).

Q3. How electromagnetic induction causes firing in a rail gun

The DC current in the rails induces a magnetic field in the gap via Ampère’s law (a steady current produces a static B‐field).

The interaction (induction) of this magnetic field with the current in the armature generates a Lorentz force that accelerates the projectile along the rails.

Thus, although no time‐varying flux is needed for the force, the underlying principle is the induction of a magnetic field by the rail currents, whose interaction with the armature current produces the propulsive force.

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a) Explain how the oscillation of the

magnet induces an EMF in the solenoid.

Induced emf is proportional to the rate of change of magnetic flux.

Ф = B.A.cos(Θ). As the magnet oscillates, B is changing, therefore emf is induced

b) Calculate the angular frequency (⍵) of

the magnet’s oscillation.

c) Determine an expression for the

induced EMF in terms of B₀ , ⍵ and other

relevant quantities.

d) Calculate the peak (maximum) induced

current in the circuit.

e) Explain what happens to the induced

EMF if the amplitude of the magnet’s

oscillation doubles.

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Self-inductance in a solenoidal electromagnet

What it is (physical idea).
Self-inductance is the property of a coil by which a changing current in the coil produces a changing magnetic flux through the same coil, which induces an emf that opposes the change (Lenz’s law). In a solenoidal electromagnet (a long coil), the coil’s own magnetic field links its turns, producing self-inductance.

When we have a current going through we will have an induced magnetic field (black line). Using faradays law, this explains the working. The current from the battery will induce the B field inside the solenoid. This than induced the current hence inducing a secondary field AKA the blue line around the solenoid

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Hand rules for electric fields and induction

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Hand rules for electric fields and induction

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Remember Lenz's law. Basically whenever something happens, the opposite will occur.

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The rapidly increasing magnetic field (into the page) induces an emf around the entire loop (Faraday’s law). That drives a single circulating current through the square. Since X and Y are identical and lie in the same series loop, they carry the same current, so they dissipate the same power and glow equally bright. (Direction check: with B increasing into the page, the induced current is counterclockwise, so both do light.)

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Remember that the cylindrical rod makes a loop with the rail. The rest of the stuff happens in there

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Review this. Seems troublesome

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In the emf formula, in the magnetic flux part, if the coil rotates at angular speed omega, it replaces the theta in BAcos(theta).

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Complete this ASAP (then ask for the worked solution)

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Complete this ASAP (then ask for the worked solution)

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A4 Rigid Body Motion Paper 2 Exam Question

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Ans: D

Remember that as the source gets closer to the observer, the frequency increases and wavelength decreases.

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1. The shift in the wavelength of light emitted by a galaxy is measured on Earth. The shift allows the determination of

A) the speed of the galaxy.

B) the velocity of the galaxy.

C) the component of the velocity along the line of sight.

D) the component of the velocity at right angles to the line of sight.

Ans: C

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2. A train approaches a stationary observer. The speed of the train is 40 m s-1 and the speed of sound is 340 m s-1. The train emits sound of frequency 5.00 x 10^2 Hz. What is the frequency heard by the observer?

A) 559 Hz

B) 567Hz

C) 447Hz

D) 441Hz

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Ans: B

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Ans: B

Because the boat measures a lower frequency (3.0 Hz) than the shore (4.0 Hz), it must be moving in the same direction as the waves (so it meets fewer crests per second). The waves are moving towards the shore, therefore the boat is also moving towards the shore.

Quantitatively (Doppler for a moving observer in a medium):

Wave speed relative to shore: v = 4 m/s

True frequency (at rest in the medium / shore): f = 4.0 Hz

Observer (boat) speed uuu in the same direction as the waves gives
fboat=f(1−u/v)

With fboat = 3.0 Hz:

3.0 = 4.0 (1 - u / 4.0) -> 3/4 = 1 - u/4 u = 1 m/s

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Ans: B

Both pulses move right toward the end at x = 5

The end is free, so a pulse reflects without inversion (the reflected pulse has the same sign as the incident one).

So the negative pulse on the right hits the free end first and reflects as a negative pulse that now travels left. It then moves toward the positive pulse coming from the left. Because the pulses have the same shape and equal but opposite amplitudes, when they meet they superpose and can cancel exactly, leaving the string momentarily flat.

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4. Waves of wavelength λ are incident on a rectangular slit of width b. In which case will the amount of diffraction through the slit be the least?

A) λ << b

B) λ < b

C) λ is approximately equal b

D) λ > b

Note: don't forget to just look at the formula

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Important Graph 1: Draw the graph for magnitude of the electric field strength (E) vs distance (r)

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Important Graph 1: Draw the graph for magnitude of the electric field strength (E) vs distance (r)

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Important Graph 2: Draw the graph for magnitude of the electric field strength (E) vs distance (r)

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Important Graph 2: Draw the graph for magnitude of the electric field strength (E) vs distance (r)

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Strings: 2 closed ends

Sound: 2 closed or 2 open

Formula: λ = 2L / n

Strings: 1 open 1 closed

Sound: 1 open 1 closed

Formula: λ = 4L / n

n must be odd

L = The length of the medium

n = The mode number or harmonic number, which is a positive integer

λ = The wavelength of the standing wave

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Q1. Ans: C (0.81c)

Explanation: use the special relativity velocity addition formula. u = -0.40c, v = 0.60c.

Q2. Ans: B

Explanation: transforming an event’s coordinates from frame S to a moving frame S’. That means using the Lorentz transformation equations. Set up the equations for movement along the x-axis, substitute the values and do the math

Q3. Ans: C [root(3) / 2 c]

Explanation: just rearrange the gamma equation

Q6. Ans: D

Explanation: In the box frame, the light just crosses the proper length (so time = length / c). In the ground frame, the box is shorter (length contraction) and the right end moves forward, so the light must chase the moving end at relative speed c−v.

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Question to understand Lorentz transformation

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Question to understand Lorentz transformation

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Ans: B

x' = γ(x - vt),

x' = [1 / sq(1 - 0.8^2)] x [1 - 0.8(0)]

x' = 5/3

t' = γ(t - vx / c^2)

ct' = γ(ct - vx / c)

ct' = 5/3 (0 - 0.8) = -4/3

Note: In relativity, space and time are treated as one combined thing called spacetime. However, you can’t directly add or compare distances (meters) with times (seconds), so time is multiplied by the speed of light to get units of distance.

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When should I use 'change in t' vs lorentz transformation 'time coordinate'

change in t: t' = gamma(t - vx/c^2)

lorentz transformation: t = t'

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Special Relativity Velocity Addition Formula

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Special Relativity Velocity Addition Formula

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u: velocity of the object in the original frame (e.g. if the original frame is earth, then the velocity of the object from earth's perspective).

v: velocity of the moving frame relative to the original (e.g. a rocket flying away from earth at 0.80c).

u′: velocity of the object in the moving frame (e.g. a satellite going at 0.85c from earth's perspective, it's gonna be different when viewed from the rocket's perspective).

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1. Relativity of Simultaneity

Example: Imagine you are standing on a train platform and two lightning bolts strike at both ends of a moving train at the same time. To you, they are simultaneous. But someone sitting in the middle of the moving train will see one flash before the other (because the train is moving toward one flash and away from the other).

Time itself depends on your frame of reference. “Now” is not universal.

2. Time Dilation

A moving clock ticks more slowly compared to a stationary clock (as seen by the stationary observer). Fast moving objects experience less time.

Example: Suppose you fly near the speed of light on a spaceship for what you think is 1 year. When you come back to Earth, people might say 10 years have passed here. You have aged less → “the twin paradox.”

3. Length Contraction

Objects moving relative to you appear shorter in the direction of motion. Space itself “squeezes” depending on motion.

Formula: L = L₀ / γ

Example: If a spaceship is 100 m long at rest, and it flies past you near the speed of light, you might measure it as only 50 m long (depending on its speed)

4. Mass Dilation (Relativistic Mass Increase)

A moving object resists acceleration more, as if it has more mass. No object with mass can reach the speed of light, because it would require infinite energy.

Formula: m = γm₀

Example: If you try to push a proton to light speed, the faster it goes, the harder it gets to accelerate. At near-light speeds, it behaves as if it’s extremely heavy.

5. Relativistic Velocity Addition

Speed of light is the fastest anything can go. Velocities don't add up at speeds of upwards of 0.2c.

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The “scaling factor” that tells you how much relativity kicks in for a given speed. Stays close to 1 for velocities up till 0.5c, then increases to infinity.

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Δt = γΔt₀

Δt0 = (proper time): the time between two events measured in the inertial frame of the moving object itself, where both events happen at the same place. Example: a muon has a lifetime of 2.2 μs when measured “onboard” with the muon’s own clock.

Δt = dilated time: the time interval measured by another observer in a different frame where the object is moving. Example: someone in the lab watching the muon sees it live longer (≈ 11 μs) because its clock is ticking slower.

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Muon Decay as proof of special relativity

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Muon Decay as proof of special relativity

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Muons live only about 2.2×10−6 s when at rest. In that time, even if moving at nearly the speed of light (0.99c), they should only travel about 0.65 km. But muons are created about 3 km up in the atmosphere, yet many are detected at Earth’s surface. How can they survive long enough?

The muon is moving at 0.99c. Due to time dilation, its lifetime is stretched:

Δt = γΔt₀, γ ≈ 7

So instead of lasting only 2.2 μs, it appears to last about 15.6 μs. In that longer time, it can travel about 4.6 km, which is enough to reach the surface.

From Earth’s point of view: muons live longer. From the muon’s point of view: the mountain is shorter.

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Ans: B

Note that the Earth observers are seeing the light reach the rocket ship, not the Earth. Read the question carefully.

Rocket's speed (v) = 0.5c (gradient of the rocket's worldline)
t' = γ(t - vx / c^2)t' = [1 / sqr(1 - 0.5^2)] [1 - (0.5c)(0.5 ly) / c^2]t' = 1.1547005(1 - 0.25) = approx. 0.87

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How the spacetime diagrams for the unboosted frame (the barn) and the boosted frame (the ladder) shows the paradox and explain how it can be resolved.

The setup:

You have a barn that is shorter than a ladder when both are at rest.

The ladder moves very fast toward the barn (say, lengthwise), with speed v close to c.

Both the barn and ladder have doors that can be opened or closed simultaneously in their own rest frame.

In the barn’s rest frame (unboosted frame):

The ladder is length-contracted because it’s moving, so it appears shorter than the barn.

Therefore, the ladder can fit entirely inside the barn for a brief instant.

The farmer can, in principle, close both doors simultaneously for a moment.

In the ladder’s rest frame (boosted frame):

The barn is moving toward the ladder.

The barn is now length-contracted, so it appears even shorter than before.

Therefore, the ladder is longer than the barn, which should be impossible.

The solution is that “simultaneous” events in one frame are not simultaneous in another.

In the barn’s frame, the two doors close at the same time.

In the ladder’s frame, the doors do not close at the same time:

The front door (where the ladder enters) closes first.

The back door (where the ladder exits) closes later.

In the space time diagram:

The barn frame (unboosted frame)

The barn’s doors are at fixed positions (x = 0 for the front and x = 10 for the back).

The ladder moves to the right (speed 0.6c), so its worldlines (front and back) are slanted.

Because the ladder moves with speed v in this frame, those worldlines have slope dx/dt = v.

The ladder frame (boosted frame)

Now the ladder is stationary (vertical worldlines)

The barn moves to the left, so its door worldlines are slanted.

The two door-closing events (same physical events as before) are not simultaneous now

The two door-closing events (same physical events as before) are not simultaneous now:

The front door closes first (t′ = −6)

The back door closes later (t′ = 0)

The barn is length-contracted to only 6.4 light-years, so the ladder is longer and cannot fit inside at one instant in its own frame.

Each frame slices spacetime differently, so what’s “simultaneous” in one is not in the other. There’s no contradiction, as both agree on the same events, just not on their timing.

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tanθ = x/ct = v/c

The tangent of the angle between the worldline and the ct axis gives the speed divided by the speed of light.

This is why it is convenient to plot ct on the vertical axis: a photon moves at speed c, and so it makes an angle of θ = tan⁻¹ (c/c) = 45° to both axes. Since nothing can exceed c, the worldline of any particle will always make an angle less than 45° with the ct axis. It is impossible for worldlines with angles greater than 45° to exist.

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In relativity, observers moving relative to one another don’t agree on:

how long something took (time dilation), and

how far apart two events were (length contraction)

Even though space and time individually depend on the observer, there’s a combined quantity that everyone agrees on — the spacetime interval.

(Δs)² = (cΔt)² - Δx²

Depending on the value of (Δs)², space-time intervals can be classified into three types:

Timelike - (Δs)2 > 0

Time separation dominates. One event can causally affect the other (a signal can travel slower than light).

Lightlike - (Δs)2 = 0

Events are connected exactly by light. (A photon could go from one to the other.)

Spacelike - (Δs)2 < 0

Space separation dominates. The events are too far apart in space for a signal to travel between them — no causal connection.

In a Minkowski diagram with axes ct (vertical) and x (horizontal). Two events are points on that diagram. The spacetime interval is like the “distance” between those two points — but measured using the rule with a minus sign.

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Whether its single or multi slit diffraction, distance doesn't affect anything other than the spacing of the fringes on the screen. Greater distance (D) = increase in the spacing of fringes on the screen (s)

s = λD / d

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7. C

1. I. Time taken same? ❌

For the observer on the ground, the light travels a longer diagonal path, so the time is longer.

(This difference in time is the basis of time dilation.)

II. Speed same? ✅

Both observers measure the same speed of light

III. Distance same? ❌

Ground observer sees the light travel a longer diagonal distance than the vertical distance measured on the train.

8. A

1. When the train slows down, the diagonal path becomes less slanted, i.e. closer to vertical.

2. Therefore:

The distance travelled by light (as seen from the ground) decreases (shorter diagonal).

The time taken also decreases, since t = distance / c and c is constant.