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Homework Help: Two optics problems

  1. Nov 8, 2007 #1
    Problem 1

    1. The problem statement, all variables and given/known data

    A goldfish's eye is 4cm from the surface of a spherical goldfish bowl of radius 10cm. Neglecting the thickness of the glass, find the apparent position and linear magnification of the eye to an observer if the refractive index of the water is 1.333. (Hint: have the light travelling from left to right to meet the surface of the bowl)

    From "Optics" 2nd edition 1996 by AH Tunnacliffe and JG Hirst. Pg 62, Question 20

    2. Relevant equations

    (1) Lreduced' - Lreduced = F

    (2) Lreduced = n/l L'reduced=n'/l'

    (3) h'/h = Lreduced/L'reduced

    (4) (n'-n)/r = F

    3. The attempt at a solution

    Using (1) and (2)
    n'/l' - 1/l = (1-1.333)/(-0.1)

    L' - 1.333/(-0.04) = 3.33
    L' = -29.95
    l' = -0.03334m

    L/L' = (1.333/-0.04)/(-29.95) = 1.11

    That gives me the right answers, but I'm really confused why that is.

    Should I do this problem with r = +0.04 or -0.04?
    Why, for L, would I treat n as 1 when it's on the side of the water?
    Why doesn't n1/n2 = real depth/apparent depth work?

    Problem 2

    1. The problem statement, all variables and given/known data
    A spherical refracting surface, separating air from glass, forms a real image twice the size of the real object. If the image is 6 times as far from F' as the object is from F', find the refractive index of the glass.

    From same source as above but Q24

    2. Relevant equations
    xx' = ff'
    h'/h = Lbar/Lbar'
    F = -n/f
    F = n'/f'
    L' - L = F

    3. The attempt at a solution

    h' = 2h

    2h/h = Lbar/Lbar'
    2 = Lbar/Lbar'

    Lbar=1/l (because in air)
    l = x+f

    Lbar' = n'/l'
    l' = 6x + f' (because image's x is 6x object's x)

    2 = Lbar/Lbar'
    2 = (1/(x+f))/(n'/(6x+f'))
    2 = (6x + f')/((x + f)n')
    2n'x + 2n'f = 6x + f'

    This is where I wonder whether the f's are equal and if not, what they are. I've found introducing r from F=(n'-n)/r (as f = -n/F and f' = n'/f') makes it even more muddled

    Last edited: Nov 8, 2007
  2. jcsd
  3. Nov 8, 2007 #2
    For your first question, you get the right answers because you did it right.
    As for the sign convention, generally the direction of the ray of incident light is taken to be positive.

    You do take n=1 for the outside because thats the refractive index of air. The bowl is kept in air, and the fish is in water, so n=1 for air and n=1.33 for water.

    n2/n1 doesnt work because that is the condition for normal viewpoint. As the rays of light arent incident normally, this doesnt work. The process is the same.

    L'/L works because thats approximately equal to the ratio of the height of the object to the height of the image. You can prove this using similar triangles. Its quite straightforward.
  4. Nov 8, 2007 #3
    For your second question, l=x-f and l'=6x+f, and unless something else is given, I think you'll have to keep the answer in terms of the radius of curvature of the refracting surface.
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