Edexcel International AS Physics Unit 2 A January 2026 Full Paper Walkthrough

Hello all. In recent comment, someone requested that go over the unit 2A, paper from 2026. So, here it is. And few things want to say in general about these papers. this is just the fourth unit two, think, I'm uploading and have uploaded four unit ones, three unit 3es and several unit fours, fives and six. And the reason for that is that in the beginning focused on unit four, five and six for no particular reason. So hopefully will gradually populate my library of videos with more past papers and I'm also working on some GCSE papers which will hopefully keep uploading and GCSE right that's the name. Yes. So yeah, mean keeping this channel alive is challenging because it's pretty much me and my sister's office. mean she provides space so can film the videos, work, etc. But it's lot of work and without funding it's really difficult to keep going. But hopefully we'll see how to handle that. So if you if you guys have any ideas please let me know in the comment section. And also please subscribe because noticed that most people just watch the videos without subscribing and that's not ideal for me and it will be great to get more subscribers. But anyway, without further ado, let's dive into this paper. Okay. And start with section The thing always say is that questions that award one mark should take you about minute. Questions that were 10 marks should take you about 10 minutes and so on and so forth. And about section sometimes you can arrive at the correct answer just by the process of eliminating the wrong answers. Okay, so that's good to keep in mind. So let's start. Which of the following units is equivalent to the volt? Okay, the way they are written is not helpful. The best way to solve this is to write down the equation. Work is charge times voltage. Right? So what are the units for work? so actually let me use the square brackets. Square brackets denotes taking the units of of quantity. So work is given in jewels. Charge is in kulum and voltage is of course in volts. So we see that voltage is just jewels per kum which is option Okay. Okay. question two. The diagram shows five energy levels in an atom. Electromagnetic radiation is incident on the atom. Which transis transition excuse me will be caused by the absorption of the lowest frequency of radiation? Okay. So, so the change in energy is equal to times the frequency. Okay. So, if frequency is low, delta is also low. Okay. And we see that the lowest such transition will be transition from four to five. So, electron in in four. So, we have an electron there and the lowest frequency radiation falls onto it. is absorbed by the electron and then it jumps the electron jumps to five. So from four to five is the correct answer. Okay. don't know why this if you noticed in the bottom of the video somewhere there the translation of something keeps appearing over and over again. don't know how to get rid of it. If you know again let me know in the comments because it's annoying. So think it tries to read my handwriting and translate it something like that. But, maybe it's setting can can fix. Let me check. So, if go here, and if go document editing, let's see. automatic screen load. screen lock, excuse me. Toolbars, zoom. don't see anything here. So, that's not it. What if we try here? No. Is there anything else here? No. Interesting. Anyway, can't figure it out. So, let me know if you know. So, let's move on to question three. Which of the following graphs shows how current varies with potential difference for negative temperature coefficient theto? So let's first discuss what the negative temperature coefficient thermister is. So device like that for device like that as increases as the voltage increases the negative temperature coefficient theister heats up. Okay. which means that the resistance falls mean decreases so current rises faster. Okay. And another thing you can write is that NTC theers are semiconductors. Okay. So if you heat them up heating makes this is what was talking about. mean it popped up in the bottom. So hitting makes electrons populate the valance space populate the valance band. Okay. So it conducts electricity better. Okay. So the correct answer is Okay. And we see that as the voltage increases the current increases. Okay. So let's move on to question number four. Okay. think figured out how to make the automatic translation go away. think had setting in the pen that use which tries to kind of correct your grammar. So switched that off. So let's see what happened. So back to question four. we have to find the Broly wavelength sorry the speed of the electron knowing that the Broly wavelength so let's start by writing lambda is over is the prince de the brol's relationship because the guy was an aristocrat okay so it's over mv so we need to solve for so is over lamb and now all we have to do is to plug in the values for 6.63 * 10 34 the mass of an electron is 9.1 sorry 11 * 1031 time lambda which is 5.47 47 * 10^ -10. Okay, which is option right and this is let me check real quick. Yes, this is an example where you can use the process of elimination because you see that in option they include the charge of the electron which is totally relevant to the velocity of the electron. Okay, same goes for option And then in option we have option is pretty much the inverse of option So that's also wrong. So three out of four are totally wrong. Okay, mean the the units do not work out because for example here we have kilograms. mean I'm talking about this option over here. Okay, we have kilograms times meters over jowls which is not speed. Okay, so so this is totally wrong. Okay, so you can use the process of elimination sometimes like in this example. All right, so think the problem with popping up of the translation thing was fixed but we will see. Question five. as temperature increases the resistance of negative temperature coefficient thermister changes. Which of the following describes the reason for this change? So you see immediately that five is related to three. mean we just talked about it and we said that negative temperature coefficient theisters are semiconductors. So if we heat them up more electrons go into the valance shell of the material. So it conducts electricity better so the resistance falls. So this is option Okay. okay. In collisions between electrons and ions occur less frequently. This is irrelevant because is more relevant to conductors. Okay. To that have ions and stuff like that and free electrons. ions in the thermister vibrate less. That's impossible. mean you heat it up. Okay, ions in the thermister vibrate more. That's option That's not the answer. So, is the correct answer because you have more conduction electrons released. So, let's see. So, the resistance reduces. Okay, let's go to question six. Now, question six and seven refer to the following information. The diagram shows circuit used to determine the internal internal excuse me resistance and of cell. Corresponding values of potential difference and current were taken after adjusting the variable resistor. graph of on the Y-axis and on the Xaxis was plotted. Which of the following is correct? Is question six. the area underneath the graph is the total energy dissipated in the cell. That's not it. The gradient of the graph is Interesting. But it's actually minus Okay. See the graph shows that and are directly proportional. They are not because the graph will look like this by the way. Okay. It so we will have and It will look like this where this is and the gradient is minus Okay. So is correct answer because the intercept of the graph is Okay. The resistance of the variable resistor is Which of the following gives the current in the cell? So in other words, we have the total resistance of the circuit as plus the intern internal resistance excuse me because they are connected in series. Therefore the current is going to be equal to the / + Okay. So that's option Okay. I'm trying to see. Okay. In seven, you can immediately eliminate eliminate options and because they ignore the resistance of the other element in the circuit. And then you can also eliminate because when you have resistances in series, you do not subtract them. Right? So again, problem seven is one of those problems in which you can arrive at the answer using the process of elimination. So let's move on to question number eight. Which of the following statements is correct for plain polarized light? So option the oscillations are in single plane with including the direction of energy transfer which is actually the correct answer. Okay, let me show you why. So polarized light means the following. First let me show you unpolarized light. So unpolarized light looks like this. is the energy transfer is in the direction of the arrow. Okay. And then you have light oscillating in all possible planes pretty much. Okay. So, let me remove this. So, something like this. Okay. That and that and so on. Okay. So this is unpolarized. Unpolarized mean you can imagine plane for example that is perpendicular to the energy transfer and contains oscillations in this direction and then you can imagine other planes containing oscillations in other directions. Now for polarized light let's say let's use black again. So let me show you what polarized light is. So so the energy transfer is in this direction. Okay. And then the polarization is perpendicular to this direction. But if you draw plane that includes this oscillations, the plane will also include the direction of energy transfer. Okay? So I'm trying to say that this black arrow or vector if you will and this vertical oscillation vector are perpendicular to each other and belong to the same plane. Okay, so is the correct answer and okay spent some time explaining it so I'm not going to read the other options and let's let's move on to question number nine. Okay, question number nine. the diagram shows uniform wire xy across which potential difference kn is applied. Okay. Which of the following correctly shows the output potential difference across So here we will use the fact that the resistance is proportional to the length of the wire. Okay. So with that logic, so out or is going to be equal to XZ this portion of the wire that has resistance related to the length of XZ over the total length XY. Okay. Over KN. Okay. So this is option option right? Yes. Okay. option for example, it's XZ over Okay, that's the wrong segment. Then over XY is option That makes no sense because it takes this portion actually. Okay. And then XY over XZ. That's that doesn't make any sense because the voltage in this case is going to be larger than not. So in this problem we can actually eliminate all wrong answers again. Okay let's move on okay to question 10. The diagram below shows standing wave on string. standing wave is created when two progressive waves traveling in opposite directions combine. Which of the following correctly indicates the phase difference between the progressive waves at and the phase difference between the progressive waves at Okay, so first of all for okay we have we have constructive interfer sorry destructive interference which means that the phase difference is pi. So we immediately exclude and and then at we have an anti-node meaning that we have constructive interference. Therefore the phase difference is 2 pi. So we eliminate and we arrive at Okay. And we and yeah mean there is nothing else to say here. So let's move on to section Okay. Question 11. the image shows satellite television dish. Okay. Electromagnetic electromagnetic radiation from communication satellite is reflected from the reflector to the detector. Okay. And we see the reflector over here. Okay. Let me erase okay the radiation used has frequency of 12.6 GHz. Calculate the wavelength of the radiation. So let me use clean page. So for question we have that lambda is simply sorry the speed of light over the frequency. Okay. So we have 3 * 10 8 over 12.6 * 10 9 which yields 0.024 which is about 2.4 4 cm. Okay. question The intensity of the radiation incident on the reflector is 4.8 * 10 to the 13 VA WS per square meter. Calculate the power of the incident radiation and the area of the satellite dish is 0.27 square meters. So let's go back again. So we know that power is equal to actually the intensity let's write it like that is power transmitted over surface area. Okay. So power is intensity time surface area or 0.27 27 squared time * 10 -3 watts per square So square meters cancel. So we get 1.3 * 10 -3 Okay. And this was question 11. don't think there is anything else to say here. Let's move on. Okay. Question 12. student investigates the resistance of the lead in pencil. pencil is used to draw rectangle of length 2 cm and width 6 On paper, creating strip of known thickness Okay. question The resistance of the strip between ends and is 55 kiloohms. Resistivity of this pencil lead is 3.5 * 105 ohm * Show that the cross-sectional area of the strip of pencil lead is about 1 * 101 square Okay, let's use clean page. So for question we will use the fact that the resistance is equal to the resistivity time the length over the surface area. So the surface area is simply equal to the resistivity time length over resistance. So this is 3.5 * 10 -5 * 2 * 10 -2 over 55,000. And the units are for resistivity we have ohms time then for the length we have meters and then for the resistance we have ohms. So ohms cancel and we are left with 1.27 27 * 101 square which is about 1 * 101 square and for those of you who are watching this video for the mean my videos for the first time whenever try to work out the problem first arrive at the final answer in algebraic form then plug in the numbers to the left okay so put all the numbers here and in in separate fraction. put all the units over there to the right. Okay? And this helps to actually make things visible, work out the units, cancel things, and so on and so forth. Okay? So, just wanted to specify for those of you who are first timers here. Okay. The pencil lead is made of mixture of graphite and clay. The pencil has 50% graphite and 50% clay. The charge carrier density for pure graphite is 3.5 * 10 24 per cubic meter. Calculate the drift velocity for the charge carriers in the pencil lead when potential difference of 6 volts is applied across the strip from to Assume the clay contributes contributes excuse me no charge carriers. Okay, so let's go back. So this question So we will pretty much use two equations. First of all, we will use the fact that current is equal to the number of charge carriers times their charge times the drift velocity times the cross-sectional area, right? yes. Okay. And we don't know the current though. So we can calculate it though using the fact that the resistance is / or Ohm's law right. So the current is equal to the voltage over the resistance. So 6 vol over 55,000 ohms yields 109 * 10 4 amps. Okay. And then we can plug into this equation. Right? So let let's go here and solve for So is what is it? Is it's the current over over * Q* Okay. And now there is trick. Think about the number of charge carriers. Are we going to be using this number as it stands or do we need to account for something? Let me know in the comments and let's see the solution. So, so first let's write the current which is 109 * 104 over actually am cramming everything into one space. So, let me delete this. Okay. And come down here. So is 109 * 10 -4. And then we have amps which which will write as kums per second. Okay, that's amps. And then we have charge carriers. I'm going to multiply by 0.5 because we have half graphite and half clay. Okay, so we have half the charge carriers. So half of 3.5 * 10 24. Okay. And then the charge and this is cubic meters in the room numerator. so let me write cleanly. So kum/ second and then cubic in the numerator. And then we also have the charge of one electron. So 1.6 6 * 1019. Okay. And we have columns down there. And then we also need the area which is run out of space. Let me move this over there. Okay. Okay. And then we need the surface area which is 1.27 27 * 10^ the -1 squar so columns cancel we have the me squar cancelling so we are left with mters/s as we should we need another clean page to write down that the drift velocity it's actually large number 3065 m/s okay and that's our answer let's read problem Okay. question Pencil leads are made with hardness from 9H, which is very hard to 9B, which is very soft. Hard pencils hard pencil leads have higher proportion of clay. Explain how the resistance of the strip will be affected if it were drawn with softer pencil. Okay. So, let's use another clean page. Okay. So, this is question 12 So, softer pencils softer pencils has more have excuse me more graphite. Okay. so higher which is the concentration of charge carriers and lower resistivity. Okay. So the resistance will be lower. Oops. The resistance will be lower. for softer pencils. Let me read it one more time to make sure that insert it correctly. okay. So yes, hard pencils have more clay. So the softer ones will have more charge carriers. Okay. So lower resistance. Okay. Question 13. Ultrasound can be used to check for cracks in metal in metal beams, excuse me. transducer emits pulse of ultrasound into metal beam. The same transducer detects the returning pulses. Part of the metal beam is shown. The beam contains three cracks and The transduc excuse me the transducer detects returning pulse from each crack. Explain why there is returning pulse from crack Okay. So let's use clean page. By the way, this unit 2A paper or papers and in general the version that's called unit 1 unit one sorry unit 2 unit 4 whatever they don't give you space to work out the problems for some reason they have it at the end but anyway it's am still trying to figure out why they have two versions now but anyway let's answer the first question so the reason Why the sound is returning from crack is the following. So, so some of the pulse so or maybe it's better to say part of the pulse part of the pulse okay transmits okay through crack because how else is it going to go to okay mean it has to go through okay sum reflects some reflects from crack Okay. So the reflection is due to change in in density or in the change of you can also say in the change is due to change in density or medium. Okay. Okay. at at the metal air interface of crack interface of crack Excuse me. think didn't explain it correctly. So in let's look at the scheme. So you send pulse. So part of it passes through crack but it's reflected to crack because you kind of change mediums right mean there is some air gap there right and therefore the sound reflects and comes back and you detect it okay so that's the explanation all right part one pulse returns 1.4 4 * 105 seconds after being emitted by the transducer. Deduce whether the pulse has returned from crack crack or crack You should take measurements from the diagram, and they give you the speed of ultrasound in metal as 5,900 per second. So, let's do this. So, the distance it covers is equal to the speed they give us. Let's call it let's call it times halves. And why is it why do we have to divide by two? Let me know in the comments. Pause. Think about it. Write it in the comments and then come back. Okay, assume you came back. The reason is because the time given is time for the pulse to go back and forth. But we only want to calculate one direction, right? Because we need to find the depth. So, so this is equal to 5,900 * 1.4 * 10 -5 / 2, which yields 0 0 413 which is about 4.1 cm. Okay, let's bring up the ruler. This is my ruler. Let's try to calculate So let's put the zero there. So for so this distance okay this distance is exactly 4 something cm. So it's crack okay hope this makes sense. So this is 4.1 cm can write in the test guess. So let's go here and say crack Okay. And believe there is part There it is. explain why ultrasound used to detect cracks cracks excuse me in metal beams usually has frequencies of megahertz rather than kilohz. Okay. So for part let's write down that the speed of of waves in general. So ultrasound in this case the speed is equal to the wavelength times the frequency right. So the wavelength is equal to the velocity over the frequency. So as sorry as the frequency increases the wavelength decreases. Now why is this important? So shorter let's write it here. Shorter lambda allows detection. Okay. Detection of smaller of smaller cracks. so so shorter lambda allows detection of smaller cracks and in general you get greater resolution. Okay. And less defraction. Okay. So that's the reason the reason actually so mean if you have small crack like this and you have smaller wavelength okay it will not defract through this crack. It will instead reflect. Okay. So that's the reason why you want smaller wavelength with these pulses. All right, if this doesn't make sense, let me know in the comments. Ask me question. I'm going to try to give better explanation. So, let's move on to the next question, which is question 14. Question 14. drivers use car headlights to emit light in dark conditions. The graph shows how current varies with the potential difference for car headlight. Okay. The headlight will emit light when it dissipates at least 35 watts of power. question determine the minimum potential difference for which the headlight will emit light. So it's little bit of weird question. it took me while to think about it, but then it appear to be easy. And we have to base our answer to what they say here. 35 Ws of power. Okay. So we have to look at the graph and find where the product * is equal to 35 And this happens over here actually. Okay. So let's go to the back of the paper and use clean page to write the answer. So at at = 9 volt is equal to 3.9 amps. And you can see it from the graph. Okay. So, let me leave this red. So, you see that that this little cross is exactly at 3.9 amps. Okay. So, so which is * is 35.1 watts. Okay. So, so the headlights start at 9 volts. Okay. So, that that's the answer. Okay. Part In car, two headlights are connected in parallel with 20 volt battery. The battery has negligible internal resistance. explain the advantages of connecting the headlights in parallel with the battery rather than in series. Okay. So So in parallel in parallel each light receives the full 12 volts. but in series it only receives 6 volts, right? it only receives 6 volts. Okay. in parallel lights operate at higher power. Okay. and are brighter because they receive again the full 12 volts. Okay. And the other advantage is that if one light fails, okay, the other still works, the other still works. And then in series, if one fails, both fail. And just had this random thought. It's kind of like physics and math. Maybe math is more maybe in math this is more profound. what mean in math sometimes you cannot jump to the next concept until you know the previous one. So it's serial knowledge. mean physics and math are serial in that for example you cannot understand anything in electromagnetism if you don't know Newton's laws for example. Okay. And the same goes with math. If you don't know derivatives you cannot learn integrals for example. So it's serial knowledge. Okay. Whereas history is kind of serial for example but not that serial. mean, you can learn about the Second World War independently of World War Okay? mean, it did play role, but if you don't know World War and you learn World War II, you're going to be fine. Something like that. So, anyway, this is very random thought. So, let's move on. okay, we are done with B1. And then we have second part. student writes the following statement. When connected to the 12vt battery, the combined resistance of two headlights in parallel is one quarter of the combined resistance of two headlight headlights, excuse me, in series. Deduce whether the student is correct. Your answer should include calculations using data from the graph. Okay, so first of all, we need to find the resistances. So let's go to the graph. So 12 volts is over here. So at 12 volts the current is 4.6 amps. Okay. So let's go back. So this part two at equals to 12 volt. Current is 4.6 amps. Okay. So is simply 12 over 4.6. 6 or 2.61 ohms. Okay. So, parallel will be 2.61 / 2 or 1.3 ohms. And the reason is because in parallel we have 1 / + 1 / is equal to 1 / total. Which means that our total is 2 / sorry / two. Right? Now at 6 volts at = to 6 vol the current is let's go see. So 6 volts is over there. So it seems to me that it's 3.2 amps. Okay. So at 6 volts the current is 3.2 amps. So the resistance is we have to we have to divide 6 over 3.2 yields 1.88 ohms. So the total resistance is going to be two times that because they are in series. So you add them up. So we get 3.75 ohms or 3.76 is better, right? okay. and let's divide we have to divide this with this and see if we get four. So 3.76 over 1.3 yields 2.9 not four. So the student is wrong. Oops. don't know why my handwriting deteriorates as as we keep going. tried to improve it but it's hard. Okay. So the student is wrong. The student is wrong. All right. So this was question 14. Let's move on to question 15. Okay. Question 15. the photograph shows musical instrument called harp or harp. don't know. it's it's an ancient Greek instrument by the way. It's what God Apollo used to play. And for those of you who are new to these videos, I'm Greek criot. And live in an island that is very kind of ancient. So there are temples actually of Apollo here in Cyprus. So anyway, irrelevant to the Let's move on. So when harp string is plucked transverse waves travel along the string forming standing wave state what is meant by transfer transverse excuse me wave. So oscillations are perpendicular to the direction of energy transfer is the answer. So let's write it let's squeeze it over here. So oscillations oscillations are perpendicular pericular to the direction to the direction of energy transfer. Okay. The string is initially made to vibrate at its lowest frequency, but the string can also be made to vibrate at higher frequencies. Add to the diagram in the answer book to show the first stationary wave on the string when it is vibrating at higher frequency. You should label any nodes and anti-nodes. Okay, so the basic frequency will be this. Okay, so this is what they try to say. But they don't want you to draw this. So let's erase this. They actually want you to draw the situation when you have one frequency higher than the basic pretty much. Okay. the mentioned this. Yeah. The first stationary wave. So this has Okay. It will have an antinode in the sorry node in the middle. So it will look like this. Okay. And then we can do something like this. So we will have node, node and node. So three nodes and we have we will have two anti-nodes. So this is the situation. Okay. Not the best lines in the world, but hope you get the idea. Okay. let's move on. Part For one string, the tension in the string is 41 Newtons and the wave travels at speed of 141 m/s. This string breaks and needs replacing with new string. The table shows the mass of four new strings, and Each new string has length of 1.5 The new string is cut is cut, excuse me, to the same length as the broken string and is placed under the same tension. that use which new string or should be used to replace the broken string. So the velocity of the waves is equal to the square root of the tension over the mass per unit length. Okay, so let's write this down. Tension over mass per unit length. Okay, because mu is / Okay. So if we take the square of the velocity, we have * length over mass. And we simply have let me erase this. We don't need it. So we have to solve for mass. So mass is tension time length over velocity squared. Let's plug in numbers. So 41 and this is newtons. length is 1.5 and then we have to divide by the velocity square 141 squared and then we have meters squared squares squared excuse me over second squared so we cancel this so we are left with * 103 and then we convert the newtons into kilogram g* over second squared and then the remaining units are second squared over So we see that everything cancels and we are left with kilograms. So this is about 3.1 Okay, which is option Okay, that's that was problem 15. by the way, this is something you should know by heart. 1 Newton is 1 kilogram * / second squar and it comes directly from is equal to mass time acceleration. So the units for are newtons. The units for mass are kilograms. The units for acceleration are m/s squared. The square brackets means that we take the units used to measure physical quantity. Okay. So this was question 15. Let's move on to question 16. Dishwashing machines use solution known as rinse aid to prevent water from leaving marks on the dishes as they dry. question student carries out an experiment to determine the refractive index of rinse aid. The photograph shows ray of light passing through rectangular container of rinse aid. Determine the refractive index of the rinse aid. you may ignore any effect from the container. So we have to measure angles now. So let's bring out okay first of all let's draw vertical line here to help us in okay not that so here this will help in measuring the relevant angles. So we basically need to measure this angle which is the incident angle and this angle which is the refractive angle. Okay, hope all this makes sense. So now we have to bring out the the ruler because it allows you to measure angles. mean you see the little tool down there. So let's let's see. Okay. So that's 90. so if we Okay. If we put it there, you see that Okay. This is what is it? Okay. the angle we want is actually 90 minus that one. Okay. Because because the ruler is measuring this angle pretty much. Okay. with respect to the to the horizontal but we need it with respect to the vertical. hope this makes sense. So the angle we need is 50. So the incident angle let me write it down. Can write? Yes, can. okay. So let's use green. So this angle is 50°. Okay. Let's measure the other one. So we have to put the ruler. excuse me. This is really difficult to do with your fingers. So, okay. All right. This is, measured it actually. So, know the answer actually, but okay. no, this is not it. What happened to it? It disappeared. okay. so this this angle is actually 34. Okay. So let's let's remove the ruler. So this angle is 34°. Okay. So but you get the idea. You have to use your the thingy. How is it called? Let me know in the comments. It's escaping my mind right now. but anyway let's let's let's do this. So let's use clean page. So we have 1 * sin theta_1 is equal to n_sub_2 * sin theta_2. So n1 is air. So 1 * sin 50 is n_sub_2 * sin 34. Okay, the two angles we were discussing. So n_sub_2 is sin 50 over sin 34. So 1.37. Okay. Part the rinse aid is stored in small reservoir in the dishwasher with an optical dipst stick to indicate the level of the liquid. The dipst stick is clear plastic cylinder with pointed pointed end that dips into the liquid show that the critical angle for light in the plastic when surrounded by rinsaid is about 60°. They give you the speed of light in rinade as 2.2 22 * 10 8 m/s and the speed of light in plastic is 2 * 10 to the 8. So let's go to part So basically of rings is 3 * 10 8 which is over the speed of light in this material. So 2.22 * 10 excuse me let's fix that. Okay. So 1.35 is the index of refraction. And for plastic we have 3 * 10 8 over 2 * 10 8. So this cancel. So we have 1.5. Okay. So plastic time sin theta critical is equal to rims time sine of 90°. And by the way total internal reflection happens when we travel from an optically denser meaning that from material that has higher index of refraction into material that has lower in index of refraction. Okay. So sine of theta critical is 1.35 over 1.5. So theta critical is 64.3° which is about 64°. Okay. let's move on to part The images below show the difference between the appearance from above of the dipst stick when the reservoir is empty and when it is full. Okay. when the reservoir is empty, the dipst stick is surrounded by air. It appears light because light from the surroundings is reflected back to the viewer. When the reservoir is full, the dipst stick is surrounded by rain shade and appears dark. the diagrams below show side views of the dipstick. Complete complete the diagrams to show the paths of light when the rings and aid level is low and when it is high. And they give you the critical angle for plastic surrounded by air as 42. 42 is the meaning of life. By the way, for those of you who know what I'm talking about, it's from book. Okay. the hitchhikers way to the galaxy something like that. So and and yeah and the story is that let me actually Google it because it now escaped my mind. mean the title of the book so the hitch let me Google it high way to galaxy. Okay. So, sorry, the title is actually the hitchhiker's guide to the galaxy and 42 in the in the story the idea is that they ask supercomputer which has artificial intelligence and everything etc. So they ask it what is the ultimate answer to the universe life and everything and after after some time when it does the calculations it outputs 42. Okay. So that's where it comes from. All right. So okay. So so we have total internal reflection here. pretty much right. And this angle is 42. So so if this angle is 42. So it's going to be the other angle here is going to be 42 as well. So 42 + 42 is 84. So in other words, I'm going to erase that mess over there. And 42 * 2 is 84 which is close to 90. So you will have something like this. It's going to travel parallel like that. And then it's going to be totally internally reflected again. So it's going to go back this way. Okay. And this is the path. Okay. And again if you draw the vertical to the plane. So this angle is 42. Okay. and so is this angle. And so are this angle. Okay, it's mess, but you get the idea. hope. If you don't, ask me in the comments. Oops, didn't mean to erase everything. Let's keep some of the information. Yeah, maybe that one. Okay, but now we go from plastic to rinse. And we just calculated by the way, why do have this page? Let's trash it. Okay. So for the other scenario, we calculated that the critical angle is 64. Okay. So we don't have we don't have total internal reflection here. And in fact we have refraction, right? So what's going to happen is that the ray will just okay it will just go through. So we will have the light ray bending away from the normal because we go from an optically denser to an optically less dense object. Okay. actually let me erase this and let's let me fix it. Okay. Maybe this is exaggerated but okay it travels like that. Okay. So I'm trying to say that this angle, the yellow one is larger than the black one. Okay. So let's call it theta is larger than theta where means incident and means refracted. Okay. All right. So this was question 16. Strange question would say but okay let's move on. Okay. question 17. The list of data formula and relationships for this paper states the following. Einstein's photoelectric equation is * frequency for photon is equal to 5 which is the work function plus 12 mv² where that's the maximum kinetic energy of the emitted electrons. Describe the photoelectric effect including an explanation of each of of the terms hf fi and 12 mv max squared in the equation. This is long writing question marked by this asterisk. and usually it's good to provide about six points that explain what's going on. So they ask us to describe. So HF is the energy of one photon. Okay. Then phi is the work function. Okay. which is the minimum energy minimum energy to remove one electron from the surface. Okay. one electron actually not necessarily from the surf from the metal is better. Okay. From the metal. okay. metal, not metal, sorry. Okay. one photon interacts interacts with one electron. mean that's the photoelectric effect. we have emission of electrons only if * is larger than or equal to 5. Okay. And 12 mv max squared is the kinetic energy max of emitted electrons. Is there anything else to mention? think not. mean this is good answer. maybe you can explain what kinetic energy max means. Which means that electrons will have energies oops energies between zero which means barely extracted from deeper into the metal. deeper into meta all the way to to key max which is surface electrons. Okay, surface electrons. All right, so this was question 17. think that's good answer. Let's move on to question 18. Okay. the speed of sound in in in seawater varies with depth as shown in diagram And we see that at the surface the speed is 1,515 m/s and all all the way down to 900 of depth. it drops to about 1,478 m/s. Okay. sound produced at depth of 900 initially travels towards the water surface. As the sound approaches depth of 700 it changes direction and follows the path shown to the point in diagram So the sound basically as they say travels this way and then it goes back down. Okay. And the question is, let's turn the page. Explain why the sound follows the path shown in diagram Your answer should refer to refraction and diagram Okay. So this is difficult question think but let's try to answer it. So this is part So speed increases. Okay. From 900 which is the lowest point, up to 700 Okay. And this is shown in diagram Okay. And you can see it. mean the speed at 900 is lower than the speed at oops 700 over there. Excuse me. Okay. All right. So that's the first thing we can write. Now another thing we can write is that the change in speed change in speed is gradual. mean it's not abrupt right. So direction. So direction changes continuously or maybe we can write changes smoothly. Okay. producing smooth curve. Producing smooth curve. And I'm talking about diagram here. diagram Okay, so waves refract away from the normal. Okay, as speed increases and the reason is that it's similar to what we just discussed with the rings earlier. So when we travel from an optically denser to an optically less dense material, the refracted light refract refracts away from the normal. It's the same case here but with sound. Okay. So let's write one more thing. So angle of incidents of incidents excuse me incidents increases increases progressively. Okay. until it reaches the critical angle. okay. And the critical angle is at 700 The critical angle is reached at 700 Okay. So total total internal reflection occurs at 700 sending the sound wave, sending the sound wave back downward. Okay. So if you draw the situation, so I'm going to draw some layers here to help you picture this. So okay, something like this. These are the layers of water. Okay. So we are at 900 Here is 800 700 and then 600 So basically we start with I'm going to exaggerate the effect little bit. So this is the sound traveling up. So here the water is dense and it goes to less dense material. So So if this is theta 1 then we're going to have it's going to bend away from the away from the normal. Okay. So the the wave that passes through is going to be like that. Okay. So now theta 2 is smaller than theta 1. Okay. And then again it's going to bend away from the normal. So it's going to bend like this. Okay. So now we're going to have an angle theta 3 here. And here we get total internal reflection and goes back down. Okay. So this is the di this is what's going on. And this is because water at 900 is denser than at 800 denser than 700 Okay. So I'm going to write it here that theta 3 is let me think excuse me actually theta 3 is larger than theta 2 and it's larger than theta 1. Okay. So we keep bending away from the normal. So every next incident angle is larger than the previous one until we reach the critical angle. So theta critical is theta 3 because we get total internal reflection. hope this makes sense. It's difficult problem. So let's move on. there's part believe. let me think. Yes, part suggest two factors that might cause the speed of sound to vary with the depth of the sea. Okay. So I'm going to I'm going to write the answer here. So temperature affects the density, right? So temperature the density of water, salinity, how much salt there is affects the speed and finally pressure. The pressure of what? Okay, this was question 18. hope the explanation makes sense. if it doesn't ask me in the comments and and I'm going to provide better explanation. So let's move on to question 19 which is the last one. Okay, question 19. laser, defraction grading and screen are set up as shown. The laser emits monochromatic light. When the laser is switched on, series of bright dots is seen on the screen. Question The diagram below shows the position of the central dot at The next bright dot appears at position The defraction grading has 450 lines per millimeter. Determine the wavelength of the light from the laser. this is question Okay. So feel like didn't read it properly, but what we have to do is basically find the angle theta. we know the two sides so it's straightforward and then we will use the defraction equation to find the wavelength. Okay. So tangent of theta this So tangent of theta is 0.5 over 1.69 which is 0.3. So theta is 16.48°. We have 450 lines per millimeter which means 4 450 over 10 to the -3 1 / right. so this means 450,000 lines per meter which means that is one over that. Okay. So is 1 / 450,000 Okay. Now sin theta is lambda but is one because this is the first the first interference bright dot. Okay. So lambda is 1 over uh,000 * sin of 16.48° 48° or lambda is 630 45 nanometers. Okay, pause the video and let me know what kind of color is this. Okay. Okay. So, actually don't know. think it's red. It's redish. 700 nanometers is red for sure. 630 is some It's probably somewhat reddish. Let me know in the comments actually what you think. So anyway, let's go to part Explain why series of bright dots is seen on the screen. So this is three mark problem which means that you have to write about three sentences explaining this. So let's use clean page. So we need the path difference. So the path difference between light between light arriving arriving to the screen from from adjacent adjacent slits. Oops. To be lambda. Okay. Or simply lambda. Okay. So waves arrive in phase at those spots. Okay, that's the other one. And then constructive constructive interference. Okay. Produces produces bright spots. That's the answer. Okay. The laser is replaced by source producing parallel beam of bright wide light suggest what would not be observed on the screen. Okay. So since we have white light we will have no isolated bright dots because we will have all wavelengths mixed up. no single color. No single color at each position. Okay. And we will have white dot white dot in the center and continuous spectra. continuous spectra on either sides. All right. And this was unit 2A from January 2026. Let me go here in the back in the end of the paper to finish this video. Okay. hope these videos are useful. It seems so. mean, you guys write wonderful comments about this video. So, thank you. but but as said, don't know for how long will be able to keep doing this. I'm currently unemployed sadly, but started looking for jobs again in academia mostly, but I'm open to freelancing opportunities, tutoring, stuff like that. like working alone in general. don't really like don't know the idea of corporate jobs but I'm open to discussing anything. So anyway these are personal thoughts. again hope these videos are useful. do read your comments guys. So try to provide stuff that you ask but okay mean it takes time. So yeah we'll see. So thanks again for watching. Please subscribe and see you in the next one. Bye.