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Refraction and Focal length

  1. Nov 9, 2007 #1

    I need to know whether Refraction changes current focal length of a camera..I have a camera focused at the bottom of a water tank(without water). After filling water to the tank i cannot see the bottom anymore focused..I need to calculate how much should I move the camera vertically to see the bottom again focused..(camera is not perpendicular to the water surface, 45°)

    I think this can be solved using Snell's law and ray tracing..but I don't know how to go on with it..

    I'm not a physics person and my high schools physics are bit rusty now..:confused:

    I'm really grateful, if somebody can tell me how to go on with this..or point out relevant literature in internet..or book..

    Thanks in advance..
  2. jcsd
  3. Nov 9, 2007 #2
    If we imagine light coming from a single point at the bottom of the tank it will change direction as it moves from the water into the air, such that the angle of refration will be less than the angle of incidence and so to see this beam we would have to raise the camera. I'm nto sure if you can work it out like this, but i will just give it a try and you can see what happens.

    Ok first we need to set up some constants :

    L = Length of container (m)
    a = Distance from edge of container to camera (m)
    h = Heigth of camera (m)
    d = Depth of water (m)

    Assuming that when the tank is emptey the camera is at height h to the bottom of the container and focused upon the centre, then we need to work out the angle at which light from the centre reaches the camera, so :

    tan(theta) = h / a+(L/2)

    Now when the container is full of water the light will travel at this angle towards the surface, so we know that when it hits the surface the angle of incidence is = to theta.

    Using snells law :

    sin(theta1)/sin(theta2) = n where n= n1/n2 and n1 = 1.333 and n2 = 1

    Rearrange to find theta2 :

    sin(theta2) = sin(theta1)/n

    Theta2 is angle that light leaves water with respect to the normal. We can now create a traingle of height of camera above water surface, camera distance from point where ray reaches surface, and theta2.


    h1 = height above surface
    b = distance of camera from point where light ray reaches surface

    b = (L/2) - (d/tan(theta)) + a

    where d is the depth of the water and theta is the first angle we calculated in this problem.

    And now,

    h1 = b/tan(theta2)

    So total height of camera above the bottom of the container will be = h + h1

    This might work, not sure though.
  4. Nov 13, 2007 #3
    Thanks for the reply..and sorry for my late response..

    I've drawn a rough sketch according to your calculations..


    The calculation with upper trangle with refraction is not clear for me..As I know, the green color light beam should go to Camera postion 1 instead of Position 2 (Because of the refraction)..So i don't get the point in considering upper trangle for new position calculation.

    Does the presence of water change the shift the focused plane? Is there any way we can go around that point..i.e to calculate the shift.

    Thanks again...
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