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Homework Help: Refractive Index of Water/Glass Combination

  1. Jan 14, 2010 #1
    If a beam of light enters water from air at 30 degrees from the normal. It then enters a glass block and exit into air. At what angle does it exit from glass to air ? My friend says it exist again at 30 degrees to normal. The refractive index of water is 1.3 and glass is 1.5.
  2. jcsd
  3. Jan 14, 2010 #2
    Your friend's forgotten Snell's Law: [tex]n_1 \sin{\theta _1}=n_2 \sin{\theta _2}[/tex]
    Where the index 1 refers to the material the light's coming from, and the index 2 refers to the material the light's entering.
  4. Jan 14, 2010 #3


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    To elaborate upon what RoyalCat says, assume you have just a block of glass (width does not matter--the pendant in me says as long as it's sufficiently thin so that not all the light is absorbed). Now assume that you have a beam of light striking at angle [itex]\theta_i[/itex]. What angle (in terms of [itex]\theta_i[/itex]) will the beam of light travel at within the glass? At what angle (again, in terms of [itex]\theta_i[/itex]) will the beam of light exit at?

    This interesting result is probably what led your friend to come up with their answer. Unfortunately, in your example, the symmetry (because there is no glass on the way in) is broken.
    Last edited: Jan 14, 2010
  5. Jan 14, 2010 #4
    For air/water/glass/water/air combination, then incoming and outgoing angle is same? This is not same for air/water/glass/air ?
  6. Jan 14, 2010 #5


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    It's the same for each individual part, so a quick sketch should convince you for the whole setup.
  7. Jan 14, 2010 #6
    its 35.23 degrees according to Mr Snell and his law :P
  8. Jan 15, 2010 #7
    For air/water surface its 30 degrees. Definetly for air/glass surface is NOT 30 degrees. Do you agree?
  9. Jan 15, 2010 #8
    Right, from water to air, the incident angle is 30, and you can figure out the refracted angle (at the water/air interface, made with the normal) using "Mr. Snell's" law.

    Now, if the water/air (it started from water, and, is entering into air) interface and the air/glass (it started from air, and, is entering into glass) interface are parallel (so that their normal's are parallel), you can figure out the incident angle for the air/glass interface ; it will be equal to the refracted angle at the water/air interface (using the simple properties of parallel lines and transversals).

    So, no, the incident angle for the air/glass interface and the water/air interface will not be the same. I think that should answer your question.

    Well, now that you've got your incident angle for the air/glass interface (it started from air, and is entering into glass), I want you to apply some logic and figure out what will be the angle of refraction when this same light ray exits glass and enters back into air?

    P.S.:- Look at the bold stuff. It points to the logic your friend is using to derive his answer. But, he's gone wrong somewhere in between
    Last edited: Jan 15, 2010
  10. Jan 15, 2010 #9
    The problem lies with the water/glass boundary. How can Mr. Snell help there? The refractive index of glass with respect to water? How to resolve that?
  11. Jan 15, 2010 #10
    Sorry, my previous answer was wrong, i didnt read the question correctly.

    Step one: n1sin1 = n2sin2 (refractive index of air = 1)

    (1)(sin30) = (1.3)(sinx)

    -> x = 22.6 degrees

    Step 2

    n1sin1 = n2sin2

    (1.3)(sin22.6) = (1.5)(siny)

    -> y=19.47degrees

    Step 3

    (1.5)sin19.47 = (1)(sinz)

    -> z = 30 degrees.

    Your friend was correct!
  12. Jan 15, 2010 #11


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    It's exactly the same law n1 sin1 = n2 sin2, you are just starting with n=1.33 and going into n=1.5
  13. Jan 16, 2010 #12
    The calculations is correct. But how to explain the physics of it?
  14. Jan 26, 2010 #13
    Anyone care to explain the physics of it?
  15. Jan 26, 2010 #14


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    In a parallel sided glass slab light is laterally shifted i.e. incident and refracted rays in the same medium are parallel to each other, whatever may be the number of different media in between medium. So the angle of incidence is equal to the angle of emergence.
  16. Jan 26, 2010 #15


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  17. Jan 26, 2010 #16
    Imagine a shopping trolly: if u push it from a smooth medium (eg marble) at an angle onto a rough concrete medium (more friction), one wheel reaches the rough medium first and thus slows down first. this causes a change of direction in which the trolley is travelling.
  18. Jan 29, 2010 #17
    "light is laterally shifted"

    What physics is that?
  19. Jan 29, 2010 #18


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    One explanation is the analogy with mechanics. Photons have energy and momentum. Their energy is E=h*f, the magnitude of the momentum is p=h/lambda, and its direction points in the direction of propagation of the light ray. You know that that momentum changes when same force acts on the particle, the change of the momentum is equal to the impulse, and is parallel to the force applied. A plane surface can act only along its normal (there is no friction). The parallel component of the momentum does not change when the photon crosses the interface between two media (or reflects from it). This parallel component is h/ lambda*sin(alfa) which is constant everywhere in a plane-parallel arrangement of layers. But lambda =v/f (f a frequency, v is the speed of light in the medium.) The refractive index of the medium is n=c/v.

    So the constancy of the parallel component of momentum involves the constancy of [tex] sin(\alpha) / \lambda = sin(\alpha) f*n/c[/tex]. As the frequency of the photon does not change we arrived at Snell's law [itex]\sin(\alpha)*n=const [/itex].

    The normal component of the momentum will change, but we know that the magnitude of the momentum is h/lambda=hf/c *n. Thus the normal component of the photon momentum is

    [tex]p_n= hf/c*\sqrt{n^2-(n_0\sin{alpha_0})^2}[/tex]

    Where n0 is the refractive index of the medium from where the light arrived and alpha0 is the angle of incidence.
    If n>n0, the normal component of the momentum increases when the light enters into the other medium while the parallel component remains the same, that is while the ligth ray encloses a smaller angle with the normal in a "denser" medium as in air.

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