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A Few Problems: MAGNETISM AND OPTICS

  1. Oct 6, 2006 #1

    app

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    INTRODUCTION: Hi, here i am stuck with a few problems. One from Magnetism and the rest from Optics.
    THE PROBLEMS: (1) A wire PQ of mass m can slide frictionlessly on two conducting rails spaced a distance l. A current i flows on into the wire PQ with the help of a current generator.There is a uniform magnetic field B perpenticular to the plane of the frame. (a)If the wire starts its motion with zero initial velocity, find its velocity after t seconds. (b) If instead, the generator is a constant voltage source of emf E and R is the total resistance of the loop, find the final velocity of the wire.
    WHAT I COULD DO:
    (1a)The first part is simple.The force on the sliding wire is given by F=Bil. Hence acceleration a=Bil/m and velocity after t seconds is v=at= Bilt/m.
    WHAT I CANNOT UNDERSTAND: In the first part, i.e. part (a), the sourse is a constant current source. Thats ok. But what is this constant voltage? What is the significance of this constant voltage? Will the velocity be uniform in the second case be uniform? If so, why? And, finally how to go about doing part (b)?
    THE PROBLEMS (Contd.): (2) (a)A transparent sphere of radius R and refractive index n is kept in air. At what distance from the surface of the sphere should a point object be placed so as to form a real image at the same distance from the sphere?
    (b) An air bubble is inside water. The refractive index of water is 4/3.At what distance from the air bubble should a point object be placed so as to form a real image at the same distance from the bubble?

    WHAT MY BOOK SAYS: (a)The answer to the first part is R/(n-1).
    (b)The answer to the second part is "The air bubble cannot form a real image".
    WHY I'M FACING A PROBLEM: Although the two questions (a) and (b) look almost the same, but i think there is a slight difference.
    THE DIFFERENCE I SEE: In both cases the object is kept in the denser medium but in part (a) the refraction takes place at a concave surface whereas in part (b) refraction takes place at the convex surface. That is what i could make out, but still i'm not quite clear about the quesion (a).
    MY EXACT PROBLEM: Although i could make out some difference but i'm not quite clear about how to go about doing the problems. I'm getting confused again and again. And why can't the air bubble in (b) form a real image? What will happen in that case then?
    THE PROBLEMS (Contd.): (3) T is a point at the bottom of a tank filled with water, as shown in the diagram. The refractive index of water is 4/3.YPT is the vertical line through T.To an observer at the position O, T will appear to be:(multiple options may be correct)
    (i)To the left of YT.
    (ii)Somewhere on YT.
    (iii)At a depth 3m below T.
    (iv)At a depth <3m below T.

    DIAGRAM IN WORDS: In case the diagram is not visible, the point O is directly above the right edge of the tank an the point T is somewhere at the middle of the bottom of the tank.
    WHAT I KNOW: The position of the apparent image will change with the angle of viewing. So, we can leave out option (ii). I think option (i) has to be correct.
    WHAT I CANNOT UNDERSTAND: But I'm cant make out what options (iii) and (iv) really want to say. Which of them is correct (if at all)? And why? I really dont understand why an image will be formed "below" T? We are viewing it from a rarer medium and the point T is in water (denser medium). Then why is the term "below T" give in the options (iii) and (iv)?
    CONCLUSION: I know i shouldn't have posted so many questions all at a time, but any help will be appreciated. Thanks a lot.
     

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    Last edited: Oct 6, 2006
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  3. Oct 6, 2006 #2

    berkeman

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    Staff: Mentor

    -1- The constant current source means that the current that generates the force is constant, even as the wire moves out on the rails. When the source is a voltage source, then the current will vary with the total circuit resistance. The resistance is related to the total path length, which is increasing as the wire slides out the rails.

    -2- You are correct that the air bubble acts like a convex lens, which cannot form a real image. You can use ray tracing to figure out the focul distance of the transparent sphere, and to show that the air bubble is a diverging lens.

    Excellent job posting your questions and thoughts, app!
     
  4. Oct 6, 2006 #3

    app

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    A Bit Confused...

    A BIT CONFUSED: ***But i'm still a bit confused about how to do the part (a) of question (2). How do i get the distance at which the object should be? The second part about the air bubble is now clear. But i'll be grateful if you could kindly explain 2(a) more clearly. ***And will the velocity of the wire be uniform in 1(b)? If so, why?
    THANKING YOU: Thanks for your help.
     
    Last edited: Oct 6, 2006
  5. Oct 6, 2006 #4

    berkeman

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    -2a- Have you learned how to do ray tracing to figure out f? There may be some other way to do it, but ray tracing is the main way I've learned. You basically have a ray that goes straight through the lens axis, and of course that ray is not diverted and goes through to the other side still straight. Then draw a parallel ray coming in from the left and a little above the first ray. When it hits the surface of the lens, it will be bent in a bit according to Snell's law. Use the angles at that point and the value of n to draw the diverted ray. Then do the same thing again as the ray leaves the lens on the right side, and extend the final ray direction until it crosses the axis ray. That is the focal distance f, where rays that come in from the left in parallel (like from an infinite object) are focused.

    Unfortunately, for a single lens of uniform n, you will get spherical aberration, so that rays that come in farther and farther off-axis will have different f values. But your problem probably wants you to use parallel rays that are only a little off-axis to calculate f, rather than giving a range of f for the distortions.

    Once you have the f of a lens, what equation would you use to calculate the image distance for a finite object distance? If you set those two distances equal in that equation, that should give you the answer for -2a-. Makes sense?


    EDIT -- BTW, a full sphere is so thick that the f may be inside the sphere itself, depending on the n. If that turns out to be the case, I'm honestly not sure how the problem should be solved.
     
  6. Oct 6, 2006 #5

    app

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    I THINK SO: Ya, i think that is the case. And i have understood the problem now.It is a sphere of radius R and not a lens.
    WHAT I HAVE UNDERSTOOD: We consider refraction at the first surface.Now if an object outside the sphere has to form an image at the same distance from the sphere, then the refraction at first surface should form a virtual image at infinity.Then only the refraction at the second surface will cause the rays to bend and form the final image at the same distance as that of the initial object.I think I'm right.But if someone finds a flaw, please let me know.
    WHAT STILL REMAINS UNANSWERED: "Will the velocity of the wire in question (1) part(b) be uniform? If so, why?" I really can't make out.Please help.
    SOME MORE PROBLEMS: I have some more optics problems which i have written out at the top of this page. Any help will be appreciated.Thanks a lot for helping.
     
  7. Oct 7, 2006 #6

    berkeman

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    Staff: Mentor

    That is very clever and insightful. I'll have to remember that.

    For new problems, it's probably best to start a new thread, especially considering the care and completeness that you use in your posts. As a tip in these kind of forums, the people who help look for certain things before opening up a thread, and if it looks like there are several replies from Homework Helpers, often we pass existing threads by looking for unanswered threads. Go ahead and start a new thread with your new questions, and by all means, please continue your exemplary style.
     
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