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    for the circle q, i got that there are 4 tagents, two of them are length 8 and two of them are length 7, anyone agree?
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    How did people do the sound waves question?
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    (Original post by DrSebWilkes)
    Look I promise myself this is the last I wade into this ... the a and b are variables in as much that they can take values that we can assign them BUT BUT BUT when they are going through the integration they are coefficients with one value; they are just values that multiply the variable.

    After that, we can "assign" a numerical value to them but that's kinda a shortcut.

    What is different though is saying "Okay, well "t" can equal ... idk ... 3 ... and then with it that equaling 3 let's integrate x ... good now let's differentiate to t"

    As I say I really can't be arsed arguing with you guys. You don't accept the basics and you don't try to explain things either.
    But the entire point is that t doesn’t have to be assigned a numerical value in order for it to be well defined. It’s kind of the foundation of algebra..
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    (Original post by kyle_19)
    for the circle q, i got that there are 4 tagents, two of them are length 8 and two of them are length 7, anyone agree?
    Yes thats correct, allthough I couldn’t finish it on the test . How many point would you get for the right equations and sketch and only working out the length 8 tangent but not giving it in the simplest terms( a root of a product of roots)?
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    Anyone taking STEP next year?
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    Here is a link explaining all the physics (although much more advanced conceptually) necessary for solving question about waves:http://www.feynmanlectures.caltech.edu/I_48.html
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    (Original post by Tomyil12345)
    Yes thats correct, allthough I couldn’t finish it on the test . How many point would you get for the right equations and sketch and only working out the length 8 tangent but not giving it in the simplest terms( a root of a product of roots)?
    Can you give me a simplification of your solution, cause it seems you made a meal out of the problem, but congrats on actually solving (unlike someone).
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    (Original post by Electric-man7)
    Can you give me a simplification of your solution, cause it seems you made a meal out of the problem, but congrats on actually solving (unlike someone).
    Do you mean how I worked out the problem?
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    (Original post by Tomyil12345)
    Do you mean how I worked out the problem?
    Yes findinng the length of the tangents.
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    i still didnt get the one on ring. How come the depth wont change? is there any critical angle?
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    and i never calculate the apparent depth, is that necessary?
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    (Original post by beeakswai)
    i still didnt get the one on ring. How come the depth wont change? is there any critical angle?
    No it doesn't involve TIR. The ring stops moving because it's hit the bottom your eye just perceived it to be shallower than it actually is because it draws a straight line coming from the refracted ray. See the attached imageName:  1510040441490264444065.jpg
Views: 42
Size:  515.7 KB
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    (Original post by joe.lai)
    No it doesn't involve TIR. The ring stops moving because it's hit the bottom your eye just perceived it to be shallower than it actually is because it draws a straight line coming from the refracted ray. See the attached imageName:  1510040441490264444065.jpg
Views: 42
Size:  515.7 KB
    but I think as the ring goes down the angle of incidence and refraction will also change? so the point on the water surface you marked out will also shift its position as the ring goes down
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    (Original post by beeakswai)
    but I think as the ring goes down the angle of incidence and refraction will also change? so the point on the water surface you marked out will also shift its position as the ring goes down
    Yes i think
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    (Original post by Tomyil12345)
    Yes i think
    how did you tackle the question?
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    2+2 is 4 minus 1 thats 3 quick maths
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    (Original post by beeakswai)
    but I think as the ring goes down the angle of incidence and refraction will also change? so the point on the water surface you marked out will also shift its position as the ring goes down
    Well yes and no. The ring isn't a point source exactly so light from it will always reach your eye,if stare at something falling in the water it doesn't appear then disappear. But this doesn't matter because the distance between the eye and the position in the water was given by the parameter w, since we were only asked to find an expression for the depth there's no reason why we can't use w in the answer
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    (Original post by joe.lai)
    Well yes and no. The ring isn't a point source exactly so light from it will always reach your eye,if stare at something falling in the water it doesn't appear then disappear. But this doesn't matter because the distance between the eye and the position in the water was given by the parameter w, since we were only asked to find an expression for the depth there's no reason why we can't use w in the answer
    However, when you let the apparent depth be constant using the straight line, you have already made the assumption that the point will never change its position, as for the ring below that depth the light ray will always pass through that point. this is not correct I suppose
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    actually i also thought about this explanation during the exam but I realised it seems doesnt work. So I changed my answer to consider the refraction between the two solutions instead and my final answer ended up having n2 inside. i am not sure whether this make sense or not
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    (Original post by beeakswai)
    However, when you let the apparent depth be constant using the straight line, you have already made the assumption that the point will never change its position, as for the ring below that depth the light ray will always pass through that point. this is not correct I suppose
    If u recall the question itself the distance w was marked on the diagram as a given parameter along with the height and the length your final answer ends up as a function of w anyways so it doesn't matter whether or not it switches position. On another note, u can draw a light ray to hit any point on the water surface and if u trace it back they all converge at the same point
 
 
 
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